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m is a three-digit integer such that when it is divided by 5, the remainder is y, and when it is divided by 7, the remainder is also y. If y is a positive integer, what is the smallest possible value of m?
Re: m is a three-digit integer such that when it is divided by
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12 Aug 2017, 13:44
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Carcass wrote:
m is a three-digit integer such that when it is divided by 5, the remainder is y, and when it is divided by 7, the remainder is also y. If y is a positive integer, what is the smallest possible value of m?
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.
Re: m is a three-digit integer such that when it is divided by
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07 Jul 2018, 12:25
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GreenlightTestPrep wrote:
Carcass wrote:
m is a three-digit integer such that when it is divided by 5, the remainder is y, and when it is divided by 7, the remainder is also y. If y is a positive integer, what is the smallest possible value of m?
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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Re: m is a three-digit integer such that when it is divided by
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07 Jul 2018, 13:24
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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.
Re: m is a three-digit integer such that when it is divided by
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26 Jul 2021, 11:30
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.