Which is the highest 3 digit number which when divided by 2, 3 and 5 successively leaves remainder of 1, 2 and 3 respectively.
Answer: 983
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Explanation :
Let us first find out the least such number which when divided by 2, 3 and 5 successively leaves remainder of 1, 2 and 3 respectively.
The least number which when divided by 5 leaves remainder of 3 is 3 itself.
Now, 3 must have been the quotient in the previous division.
∴ The number which was divided by 3 leaving a remainder of 2 = 3 × 3 + 2 = 11
Now, 11 must have been the quotient in the previous division.
∴ The number which was divided by 2 leaving a remainder of 1 = 2 × 11 + 1 = 23
∴ The least number which when divided by 2, 3 and 5 successively leaves remainder of 1, 2 and 3 respectively is 23.
∴ A number which when divided by 2, 3 and 5 successively leaves remainder of 1, 2 and 3 respectively = 2 × 3 × 5 × k + 23 = 30k + 23
Now, 30k + 23 < 1000
⇒ 30k < 977
⇒ highest possible value of k = 32
⇒ Highest such three digit number = 30 × 32 + 23 = 983
Hence, 983.
Solution
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