Question

There is such a positive integer: if it is divided by 5, the remainder is 3; if it is divided by 6, the remainder is 4; if it is divided by 7, the remainder is 1. Find the least possible value of such number.

Answer 1

• The least possible number is 148

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Let the number be N.

If N is divided by 5 the remainder is 3:

• N = 5k + 3,

Similarly:

• N = 6p + 4,

• N = 7q + 1.

We can observe N + 2 is divisible by both 5 and 6, then it can be represented as:

• N + 2 = 30m,

so

• 30m = 7q + 1 + 2 or
• 30m - 3 = 7q

The left side is divisible by 3, then q = 3t:

• 30m - 3 = 7*3t
• 10m - 1 = 7t

The least m and t would be m = 5 and t = 7.

Then:

• N + 2 = 30*5 = 150 ⇒ N = 148 is the least possible number.

Proof:

• 148/5 = 29 (rem 3)
• 148/6 = 24 (rem 4)
• 148/7 = 21 (rem 1)