Algebra - Number Theory - Previous Year CAT/MBA Questions
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The least common multiple of a number and 990 is 6930. The greatest common divisor of that number and 550 is 110. What is the sum of the digits of the least possible value of that number?
- (a)
6
- (b)
9
- (c)
14
- (d)
18
- (e)
None of the remaining options is correct
Answer: Option C
Text Explanation :
Workspace:
Consider a 4-digit number of the form abbb, i.e., the first digit is a (a > 0) and the last three digits are all b. Which of the following conditions is both NECESSARY and SUFFICIENT to ensure that the 4-digit number is divisible by a?
- (a)
b is divisible by a
- (b)
b is equal to 0
- (c)
21b is divisible by a
- (d)
9b is divisible by a
- (e)
3b is divisible by a
Answer: Option E
Text Explanation :
Workspace:
Let x and y be two positive integers and p be a prime number. If x(x – p) – y(y + p) = 7p, what will be the minimum value of x – y?
- (a)
1
- (b)
3
- (c)
5
- (d)
7
- (e)
None of the above
Answer: Option E
Text Explanation :
Given, x(x – p) – y(y + p) = 7p
⇒ x2 - xp - y2 - yp = 7p
⇒ x2 - y2 - xp - yp = 7p
⇒ (x - y)(x + y) - p(x + y) = 7p
⇒ (x + y)(x - y - p) = 7p
Since, 7 and p are both prime numbers, we have
⇒ (x + y)(x - y - p) = 7 × p or 7p × 1
i.e., one (x + y) and (x - y - p) can be 7 and p or 7p and 1.
Case 1: one (x + y) and (x - y - p) can be 7 and p
⇒ (x + y) + (x - y - p) = 7 + p
⇒ 2x = 7 + 2p
⇒ x = 3.5 + 2p
This is not possible as x should be a integer.
Case 2: one (x + y) and (x - y - p) can be 7p and 1.
⇒ (x + y) + (x - y - p) = 7p + 1
⇒ 2x = 8p + 1
⇒ x = 4p + 0.5
This is not possible as x should be a integer.
∴ No value of integral value of x is possible.
Hence, option (e).
Workspace:
A supplier receives orders from 5 different buyers. Each buyer places their order only on a Monday. The first buyer places the order after every 2 weeks, the second buyer, after every 6 weeks, the third buyer, after every 8 weeks, the fourth buyer, every 4 weeks, and the fifth buyer, after every 3 weeks. It is known that on January 1st, which was a Monday, each of these five buyers placed an order with the supplier.
On how many occasions, in the same year, will these buyers place their orders together excluding the order placed on January 1st?
- (a)
1
- (b)
5
- (c)
2
- (d)
4
- (e)
3
Answer: Option C
Text Explanation :
The supplier receives his orders from the five buyers once every 2 weeks, once every 6 weeks, once every 8 weeks, once every 4 weeks, and once every 3 weeks.
The number of occasions where all the five buyers place the order on the same day is :
The LCM of the 5-time frames during which the 5 buyers place their orders :
Hence the LCM is :
(2, 6, 8, 4, 3).
= 24 weeks.
Once every 24 weeks, all five of them place the order simultaneously.
A year has 53 weeks in total :
Hence all five of them place the orders after 24 weeks, 48 weeks.
Workspace:
The sum of the cubes of two numbers is 128, while the sum of the reciprocals oftheir cubes is 2. What is the product of the squares of the numbers?
- (a)
64
- (b)
256
- (c)
16
- (d)
48
- (e)
32
Answer: Option C
Text Explanation :
Considering the two numbers to a, b :
We were given that:
a3 + b3 = 128
+ = 2
= 2 =
k = 64.
Hence a3 . b3 = 64
a*b = 4 and a2 .b2 = 16
Workspace:
Nadeem’s age is a two-digit number X, squaring which yields a three-digit number,whose last digit is Y. Consider the statements below:
Statement I: Y is a prime number
Statement II: Y is one-third of X
To determine Nadeem’s age uniquely:
- (a)
either of I and II, by itself, is sufficient.
- (b)
only II is sufficient, but I is not.
- (c)
only I is sufficient, but II is not.
- (d)
it is necessary and sufficient to take I and II together.
- (e)
even taking I and II together is not sufficient.
Answer: Option D
Text Explanation :
The age of Nadeem is a two-digit number. When squared yields a three-digit number whose last digit is Y. Y is a prime number.
Using statement 1:
When a number is squared :
The last digit of the number can be :
1, 2, 3, 4, 5, 6, 7, 8, 9. When squared the last digit can be :
For a number ending with 1: 1
For a number ending with 2: 4
For a number ending with 3: 9
For a number ending with 4: 6
For a number ending with 5: 5
For a number ending with 6: 6
For a number ending with 7: 9
For a number ending with 8: 4
For a number ending with 9: 1
The only possible prime number is 5.
Hence the last digit of X is 5 and Y is 5.
Using statement 2 : Y = X/3.
This alone cannot be sufficient to determine the possibilities for Y and X.
Combining both the statements :
Since Y = 5, then the value of X = 15.
The age is equal to 15.
Workspace:
Wilma, Xavier, Yaska and Zakir are four young friends, who have a passion for integers. One day, each of them selects one integer and writes it on a wall. The writing on the wall shows that Xavier and Zakir picked positive integers, Yaska picked a negative one, while Wilma’s integer is either negative, zero or positive. If their integers are denoted by the first letters of their respective names, the following is true:
W4 + X3 + Y2 + Z ≤ 4
X3 + Z ≥ 2
W4 + Y2 ≤ 2
Y2 + Z ≥ 3
Given the above, which of these can W2 + X2 + Y2 + Z2 possibly evaluate to?
- (a)
9
- (b)
0
- (c)
4
- (d)
6
- (e)
1
Answer: Option D
Text Explanation :
Given that X, Z are positive Y is negative and W can be either positive or zero or negative.
The given conditions are:
W4 + X3 + Y2 + Z ≤ 4
X3 + Z ≥ 2
W4 + Y2 ≤ 2
Y2 + Z ≥ 3
For W4 + Y2 ≤ 2. Since Y is negative but Y2 is always positive and must be less than 2 because W4 is a nonnegative value. Hence Y = -1 is the only possibility. For W this can take any value among -1, 0, 1.
Y2 + Z ≥ 3. Since Y = -1, Z must be at least equal to 2 so the value of Y2 + Z ≥ 3 is greater than 2.
X is a positive value and must at least be equal to 1.
The condition: W2 + X2 + Y2 + Z2 here has all the independent values:
X2, Y2, Z2, W2 are non-negative.
W4 + X3 + Y2 + Z ≤ 4:
Since the value of Z is at least equal to 2 the value of Y2 is equal to 1.
Since X is a positive number in order to have the condition of W4 + X3 + Y2 + Z ≤ 4 satisfied. The value of Z must be the minimum possible so that X3 + Y2 + Z to have a value equal to 4 when X takes the minimum possible positive value equal to 1.
Hence X must be 1. W must be equal to 0 so that:
W4 + X3 + Y2 + Z ≤ 4. The sum = (0 + 1 + 1 + 2) = 4. The only possible case.
The value of W2 + X2 + Y2 + Z2 = (0 + 1 + 1 + 4) = 6.
Workspace:
Fatima found that the profit earned by the Bala dosa stall today is a three-digit number. She also noticed that the middle digit is half of the leftmost digit, while the rightmost digit is three times the middle digit. She then randomly interchanged the digits and obtained a different number. This number was more than the original number by 198.
What was the middle digit of the profit amount?
- (a)
1
- (b)
2
- (c)
6
- (d)
This cannot be solved with only the given information
- (e)
8
Answer: Option B
Text Explanation :
From the given conditions :
Considering the three-digit number to be a b c.
With the given conditions :
a = 2b, c = 3b.
Hence the number is of the form : 2b b 3b.
Since all three of the values must be less than 10 and non-negative:
This takes values : b = 1, b = 2, b = 3.
Hence the possible numbers are : (213, 426, 639) :
The interchanged number must be greater than the original by 198.
Hence the different rearrangements for the three numbers are :
213 : (312, 321, 132, 123, 231).
426 : (462, 624, 642, 246, 264)
639 : ( 693, 963, 936, 396, 369)
The only possible value which is higher than the original by 198 is :
(426, 624).
The middle digit is 2.
Workspace:
The Madhura Fruits Company is packing four types of fruits into boxes. There are 126 oranges, 162 apples, 198 guavas and 306 pears. The fruits must be packed in such a way that a given box must have only one type of fruit and must contain thesame number of fruit units as any other box.
What is the minimum number of boxes that must be used?
- (a)
21
- (b)
18
- (c)
44
- (d)
42
- (e)
36
Answer: Option C
Text Explanation :
The number of oranges, apples, guavas, and pears = 126, 162, 198, and 306.
Each box must contain an equal number of fruits with only one type of fruit. The additional condition provided is that there should be a minimum number of boxes in total.
The distribution is possible in multiple ways in such a way that distribution in each box is placed in such that each box contains a certain number of fruits n which is a factor for all the four given number of fruits :
Arrangement of 1 fruit of one kind in a basket.
2 is a factor of 126, 162, 198, and 306. So we can place 2 fruits of a particular kind in a basket.
Since we were asked for the minimum number of boxes this is possible when a maximum number of fruits of a kind are placed in a box.
Hence each box must contain the Highest common factor for the four numbers :
The prime factorization for the four numbers:
126 : 2 ∙ 7 ∙ 9, 162 : 2 ∙ 9 ∙ 9, 198 : 2 ∙ 9 ∙ 11, 306 = 2 ∙ 9 ∙ 17
The HCF is 18.
The number of boxes required for each :
, , ,
7 + 9 + 11 + 17 = 44.
Workspace:
What is the remainder if 1920 – 2019 is divided by 7?
- (a)
5
- (b)
1
- (c)
6
- (d)
0
- (e)
3
Answer: Option A
Text Explanation :
= = = = = = = 4
= = -1
= 4 - (-1) = 5
Hence, option (a).
Workspace:
When expressed in a decimal form, which of the following numbers will be nonterminating as well as non-repeating?
- (a)
- (b)
sin21° + sin22° + … + sin289°
- (c)
- (d)
- (e)
Answer: Option E
Text Explanation :
We need to check options.
Option (a):
= = = 0.5 (terminating)
Option (b): sin21° + sin22° + … + sin288° + sin289°
= sin21° + sin22° + … + sin245° + … + cos22° + cos21°
= 44 + ½ = 44.5 (terminating)
Option (c):
= = 3 + 22/7 (terminating)
Option (d):
= = 4 (non-terminating)
Option (e):
= 6 - 4 + √3 = 2 + √3 (non-terminating and non-repeating)
Hence, option (e).
Workspace:
Consider the four variables A, B, C and D and a function Z of these variables, Z = 15A2 − 3B4 + C + 0.5D. It is given that A, B, C and D must be non-negative integers and that all of the following relationships must hold:
i) 2A + B ≤ 2
ii) 4A + 2B + C ≤ 12
iii) 3A + 4B + D ≤ 15
If Z needs to be maximised, then what value must D take?
- (a)
15
- (b)
12
- (c)
0
- (d)
10
- (e)
5
Answer: Option B
Text Explanation :
To maximise Z, we need to maximise A, C and D and minimise B.
Given, 2A + B ≤ 2
A can be either 1 or 0.
Case 1: If A = 1, minimum value of B = 0
From (ii): Maximum value of C = 8
From (iii): Maximum value of D = 12
⇒ Z = 15A2 − 3B4 + C + 0.5D = 15 – 0 + 8 + 6 = 29
Case 2: If A = 0, minimum value of B = 0
From (ii): Maximum value of C = 12
From (iii): Maximum value of D = 15
⇒ Z = 15A2 − 3B4 + C + 0.5D = 0 – 0 + 12 + 7.5 = 19.5
We can see that maximum value of Z = 29, when D = 12.
Hence, option (b).
Workspace:
When opening his fruit shop for the day a shopkeeper found that his stock of apples could be perfectly arranged in a complete triangular array: that is, every row with one apple more than the row immediately above, going all the way up ending with a single apple at the top.
During any sales transaction, apples are always picked from the uppermost row, and going below only when that row is exhausted. When one customer walked in the middle of the day she found an incomplete array in display having 126 apples totally. How many rows of apples (complete and incomplete) were seen by this customer? (Assume that the initial stock did not exceed 150 apples.)
- (a)
14
- (b)
13
- (c)
15
- (d)
11
- (e)
12
Answer: Option E
Text Explanation :
In the given triangular array, 1st row will have 1 apple, 2nd row will have 2 apples and so on.
∴ nth row will have n apples.
⇒ Total number of apples initially = 1 + 2 + 3 + … + n =
Initially total number of apples should be greater than 126 but less than or equal to 150
∴ 126 < ≤ 150
Only integral value of n satisfying this is 16.
⇒ = 136
∴ Initially there are 16 rows and 136 apples arranged in a triangular array in these 16 rows.
As the customer observes 126 apples, it would mean that 10 apples were removed.
Now, apples are removed starting from Row 1, which has 1 apple. So apples must have been removed in such a way with 1 apple from Row 1, 2 apples from Row 2, 3 apples from Row 3 and 4 apples from Row 4.
∴ Total 10 apples were removed the first 4 rows.
As out of 16 rows, apples are removed till Row 4, it would mean that only 16 - 4 or 12 rows of apples are visible to the customer.
Hence, option (e).
Workspace:
If x2 + x + 1 = 0, then x2018 + x2019 equals which of the following:
- (a)
-x
- (b)
x
- (c)
None of the others
- (d)
x - 1
- (e)
x + 1
Answer: Option A
Text Explanation :
x2 + x + 1 = 0 …(1)
⇒ x2 = - x – 1 …(2)
Multiplying x in (2)
⇒ x3 = - x2 – x
⇒ x3 = 1 …(3) [From (1)]
Now,
x2018 + x2019 = x2018(1 + x)
= x2018 × -x2
= -x2020
= - (x3)673 × x
= - 1 × x
= - x
∴ x2018 + x2019 = - x
Hence, option (a).
Workspace:
We have two unknown positive integers m and n, whose product is less than 100.
There are two additional statement of facts available:
- mn is divisible by six consecutive integers { j, j + 1,...,j + 5 }
- m + n is a perfect square.
Which of the two statements above, alone or in combination shall be sufficient to determine the numbers m and n?
- (a)
Statements 1 and 2 together are not sufficient, and additional data is needed to answer the question.
- (b)
Both statements taken together are sufficient to answer the question, but neither statement alone is sufficient.
- (c)
Each statement alone is sufficient to answer the question.
- (d)
Statement 2 alone is sufficient, but statement 1 alone is not sufficient to answer the question.
- (e)
Statement 1 alone is sufficient, but statement 2 alone is not sufficient to answer the question.
Answer: Option B
Text Explanation :
Looking at the initial statement , we know that mn < 100.
Looking at statement I ,we get to know that the product mn has to be a multiple of 10 since it is divisible by 6 consecutive integers .So the product can be either 10, 20, 30 ….90
Now of all these numbers only 60 is divisible by 6 consecutive numbers i.e. numbers 1 to 6.60 can be expressed as a product of 2 nos. in the following ways : 1 × 60, 2 × 30, 3 × 20, 4 × 15, 5 × 12, 6 ×10
So from statement I alone we cannot determine values of m and n. Looking at statement II alone determine values of m and n as the only information provided to us is that “m + n” is a perfect square. So we can have numerous possibilities for m and n [e.g (7, 9), (2, 7), (1, 3) etc ]
Combining both statements out of (1, 60), (2, 30), (3, 20), (4,15), (5, 12), (6, 10), the only pair of values such that “m+n” is a perfect square is (6, 10). Hence both statements are required to answer the question.
Hence, option (b).
Workspace:
Find the value of the expression: 10 + 103 + 106 + 109
- (a)
1010101010
- (b)
1001000010
- (c)
1001000110
- (d)
1001001010
- (e)
100010001010
Answer: Option D
Text Explanation :
10 + 103 + 106 + 109 =
10 +
1000 +
1000000 +
1000000000 +
= 1001001010
Hence, option (d).
Workspace:
X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers (408X)63 and (789Y)85 are the same. What will be the possible value(s) of (X + Y)?
Example: If M = 3 then the digit at unit's place of the number (2M) is 3 (as the number is 23) and the digit at unit's place of the number (2M)2 is 9 (as 232 is 529).
- (a)
9
- (b)
10
- (c)
11
- (d)
12
- (e)
None of the above
Answer: Option B
Text Explanation :
For various powers the units digit of a number always a cycle of 4 terms.
We need unit’s digits of (X)63 and (Y)85 to be same
Unit’s digit of X63 will be same as unit’s digit of X3
Similarly, unit’s digit of Y85 is same as unit’s digit of Y1.
∴ unit’s digits of (X)3 should be same as units digit of Y.
This is possible (from the table) when (X, Y) = (2, 8) or (8, 2) or (7, 3) or (3, 7).
In any of these cases X + Y = 10.
Hence, option (b).
Workspace:
David has an interesting habit of spending money. He spends exactly £X on the Xth day of a month. For example, he spends exactly £5 on the 5th of any month. On a few days in a year, David noticed that his cumulative spending during the last 'four consecutive days' can be expressed as 2N where N is a natural number. What can be the possible value(s) of N?
- (a)
5
- (b)
6
- (c)
7
- (d)
8
- (e)
N can have more than one value
Answer: Option B
Text Explanation :
Sum of any four consecutive numbers = x + x + 1 + x + 2 + x + 3 = 4x + 3.
This sum is never divisible by 4. But 2N is always divisible by 4 (for N ≥ 2).
∴ The four consecutive days do not fall in the same month. Let’s look at the various possibilities.
Now, there are 4 possibilities for number of days in a month.
The only possibility for cumulative spending during the last 'four consecutive days' to be expressed as 2N = 64 = 26.
∴ N = 6
Hence, option (b).
Workspace:
An institute has 5 departments and each department has 50 students. If students are picked up randomly from all 5 departments to form a committee, what should be the minimum number of students in the committee so that at least one department should have representation of minimum 5 students?
- (a)
11
- (b)
15
- (c)
21
- (d)
41
- (e)
None of the above
Answer: Option C
Text Explanation :
The five departments of each department has 50 students.
We have to minimise the number of students in the committee so that at least one department should have representation of minimum 5 students
Let us maximise the number of students in the committee so that no department has representation of 5 students. i.e., any department can have maximum 4 students.
∴ If we select 4 students from each department i.e. a total of 20 students, still no department will have representation of 5 students.
Now if we select one more student, we will have representation of 5 students from at least 1 department.
∴ Minimum number of students in the committee so that at least one department should have representation of minimum 5 students is 21.
Hence, option (c).
Workspace:
If N = (11p+7)(7q-2)(5r+1)(3s) is a perfect cube, where p, q, r and s are positive integers, then the smallest value of p + q + r + s is:
- (a)
5
- (b)
6
- (c)
7
- (d)
8
- (e)
9
Answer: Option E
Text Explanation :
In order for N to be a perfect cube, every prime number should be a power of multiple of 3 i.e., 0, 3 , 9 …
Also, we need to consider minimum possible values of p, q, r and s. (p, q, r and s > 0)
For 3s to be a perfect cube, minimum value of s = 3.
For 5r+1 to be a perfect cube, minimum value of r = 2.
For 7q-2 to be a perfect cube, minimum value of q = 2.
For 11p+7 to be a perfect cube, minimum value of p = 2.
So, smallest value of p + q + r + s = 2 + 2 + 2 + 3 = 9
Hence, option (e).
Workspace:
For two positive integers a and b, if (a + b)(a+b) is divisible by 500, then the least possible value of a × b is:
- (a)
8
- (b)
9
- (c)
10
- (d)
12
- (e)
None of the above
Answer: Option B
Text Explanation :
500 = 22 × 53
For (a + b)(a+b) to be divisible by 500, it should have power of 2 as well as 5 in it.
∴ a + b should have power of both 2 as well as 5.
The least such number is 1010.
∴ a + b = 10.
∴ a × b is minimum when a = 9 and b = 1 (or vice-versa)
Hence, the least possible value is 9.
Hence, option (b).
Workspace:
If a, b and c are 3 consecutive integers between –10 to +10 (both inclusive), how many integer values are possible for the expression ?
- (a)
0
- (b)
1
- (c)
2
- (d)
3
- (e)
4
Answer: Option C
Text Explanation :
If a, b and c are three consecutive numbers
⇒ a = b – 1 and c = b + 1
Substituting this in the expression we get the expression we get,
=
=
=
Only for b = ±1 is the above expression an integer.
Hence, option (c).
Workspace:
An ascending series of numbers satisfies the following conditions:
- When divided by 3, 4, 5 or 6, the numbers leave a remainder of 2.
- When divided by 11, the numbers leave no remainder.
The 6th number in this series will be:
- (a)
242
- (b)
2882
- (c)
3542
- (d)
4202
- (e)
None of the above
Answer: Option C
Text Explanation :
LCM (3, 4, 5, 6) = 60
From (i), numbers are of the form 60k + 2.
From (ii), numbers are of the form 11m.
11m = 60k + 2 = 55k + (5k + 2)
11 divides (5k + 2)
Units digit of (5k + 2) is 2 or 7
So, values of (5k + 2) are 11 × 2, 11 × 7, 11 × 12, 11 × 17, 11 × 22, 11 × 27, … and so on.
We need to find the 6th number of the ascending series.
∴ 5k + 2 = 11 × 27 ⇒ k = 59
The required number = 60 × 59 + 2 = 3542
Hence, option (c).
Workspace:
If the last 6 digits of [(M)! – (N)!] are 999000, which of the following option is not possible for (M) × (M – N)?
Both (M) and (N) are positive integers and M > N. (M)! is factorial M.
- (a)
150
- (b)
180
- (c)
200
- (d)
225
- (e)
234
Answer: Option B
Text Explanation :
M and N are positive integers such that M > N
∴ M! – N! = abc…999000
∴ [M(M − 1)(M − 2)……N!] – N! = abc…999000
∴ N!{[M(M − 1) (M − 2)……] − 1} = abc…999000
Let the term in the square bracket be x.
Since M is a positive integer, the term in the square brackets i.e., x is also a positive integer, and hence, (x – 1) is also a positive integer.
∴ N!(x – 1) = abc…999000
Hence, the maximum number of zeroes in N! is 3.
∴ N! ≤ 19 (because from 20! onwards, each factorial has atleast 4 zeroes)
Now, there are 4 possible ranges for N:
- N = 0 to 4 (no zeroes in N!)
- N = 5 to 9 (1 zero in N!)
- N = 10 to 14 (2 zeroes in N!)
- N = 15 to 19 (3 zeroes in N!)
Consider case 3, where there are 2 zeroes in N!
Since N!(x – 1) = abc…999000, the third zero on the LHS should come from (x – 1).
For this, x has to be of the form pqrs…1 i.e., x has to be an odd number.
Now, there are two possibilities:
1. M = N + 1
Here, M! – N! = (M × N!) – N! = N!(M – 1)
In this case, M = x
Since N = 10 to 14 and M – 1 should end in 0, M = 11 and N = 10
Hence, M(M – N) = 11(11 – 10) = 11
This does not tally with any of the options.
Hence, M ≠ N + 1
2. There is at least one integer between M and N
Hence, x comprises a series of consecutive integers (at least two as explained above) multiplied with each other. So, there has to be at least one even number in this series.
Hence, x can never be odd.
Hence, the third zero on the LHS can never come from (x – 1).
Hence, there cannot be 2 zeroes in N!. Similarly, it can be proved that there cannot be 1 zero or no zeroes in N!.
Hence, N! has three zeroes i.e. 15 ≤ N ≤ 19
Now, M(M – N) = M2 – NM can take four of the five given values.
Hence, there are five possible equations – one for each option.
Consider option 1: M2 – NM – 150 = 0
Here, for N = 15 to 19, see if there exists a value of M (> N) that gives positive integral roots in this equation.
For N = 19, the equation becomes
M2 – 19M – 150 = 0 i.e. M = 25 or M = −6. Here, M = 25 is valid.
Similarly, consider each option.
Option 3:
For N = 17, the equation becomes
M2 – 17M – 200 = 0 i.e. M = 25 or M = −8. Here, M = 25 is valid.
Option 4:
For N = 16, the equation becomes
M2 – 16M – 225 = 0 i.e. M = 25 or M = −9. Here, M = 25 is valid.
Option 5:
For N = 17, the equation becomes
M2 – 17M – 234 = 0 i.e. M = 26 or M = −9. Here, M = 26 is valid.
However, no value of N from 15 to 19 gives an integral solution for M in M2 – NM – 180 = 0
Hence, M(M – N) can never be 180.
Hence, option (b).
Note: There is a logical flaw in this question as the difference of two factorials can never end in 999000. The actual question should have had the last six digits as abc000.
Workspace:
A three-digit number has digits in strictly descending order and divisible by 10. By changing the places of the digits a new three-digit number is constructed in such a way that the new number is divisible by 10. The difference between the original number and the new number is divisible by 40. How many numbers will satisfy all these conditions?
- (a)
5
- (b)
6
- (c)
7
- (d)
8
- (e)
None of the above
Answer: Option B
Text Explanation :
Units digit of the number must be 0.
Let 100x + 10y is the number such that x > y.
New number obtained by changing the digits is also divisible by 10.
So, only x and y are to be interchanged
∴ New number is of the form = 100y + 10x
Difference = 90x – 90y = 90(x – y)
For the difference to be divisible by 4,
(x – y) has to be divisible by 4.
(x – y) = 4 or 8
So, y = 1 to 5
For y = 1 to 5, x = (1+4) to (5+4) i.e., 5 to 9
One more possibility for y = 1 is x = 9.
Thus, in all 6 numbers satisfy the given conditions.
Hence, option (b).
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