There's a nice rule that say, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3

When m is divided by 5, the remainder is y
So, m = 5k + y for some integer k

When m is divided by 7, the remainder is y
So, m = 7j + y for some integer j

Since both equations are set to equal m, we can write: 5k + y = 7j + y
Subtract y from both sides to get: 5k = 7j
Well, 5k represents a multiple of 5, and 7j represents a multiple of 7
So, what's the smallest 3-digit number that is a multiple of 5 AND a multiple of 7?

The smallest 3-digit number is 100
100 is a multiple of 5, but it's NOT a multiple of 7

Next we have 105
105 is a multiple of 5, AND it's a multiple of 7
Now be careful. This does NOT mean that m = 105

When we divide 105 by 5 we get a remainder of 0, but we're told that the remainder (y) is a POSITIVE INTEGER.
To MINIMIZE the value of m, we need a super small remainder.
The smallest possible non-zero remainder is 1.
105 + 1 = 106

So, 106 is the smallest possible 3-digit value of m.

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How could i know which numbers i should use in general? What is your strategy of restricting the range of possible values for this problem?

This question is quite tricky.

However, 99% of the time, the more a tricky question is, the more there is always a shortcut or at a closer inspection a solution is suddenly behind the curve.

Reading carefully the stem, it says that you have to consider a 3 digit integer number and you have to find the least possible value.

So, a 3 digit number at least is 100. Now, you do also know that when it is divided by 5 y is the remainder and when it is divided by 7 is yet y the remainder, which means is the same number.

A common number that divides evenly 7 and 5 is 105. 105 divided by 5 AND 7 has no rest or reminder. From this is easy to think that a number divided by both 5 and 7 with the same reminder y, for instance, the reminder is 1, is 106.

106 is the least number with 3 digits you can have when you divide it by 5 and 7.

Hope this helps

We have m=5a+y=7b+y where 0<y<5 -> m=35c+y where 0<y<5 (hence min y = 1).

Note that 35*3=105 -> m=35*3+1=106.
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LCM of 5 and 7 is 35.
The least possible 3 digit number is 105.

Then the only possible smallest possible value of m is 106, which leave the same remainder when divided by 5 and 7.

Hence 106 is the answer

The way to solve this problem is to realize that if you solve for the least common multiple of 5 and 7 that has 3 digits, it leaves you with 105. If you divide 105 by 5 and 7 your remainder is 0. If you add one to 105 = 106, then your remainder is 1 for both values.