When an integer n is divided by 24, the remainder is 21. Which of the following must be a divisor of n?
A) 3
B) 4
C) 5
D) 7
I am solving it like if n is divided by 24( 2^3, 3) and giving a remainder of 21(7,3) then it must have at least 2^3=8 in it that’s why they are not appearing in remainder. And with this logic I am choosing B. But the correct option is A.
Hi,
Your logic does not work here as the integer n is not actually divisible by 24. I think choosing numbers would work great here. Find the numbers that when divided by 24 would give 21 as the remainder.
21, 45, 69, 93, …
The only option which is a divisor of all of them is A(3).
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Yup I understand the numbers approach. But I was curious why my logic is not working here. If you can point error in my logic?
If I’m understanding it correctly then you’re trying to solve the equation with 21 as a quotient and not a remainder.
This logic is broken. What you can do is say that n + 3 divides 24 (call its quotient k). Then,
\frac{n + 3}{24} = k
n + 3 = 24k \\
n = 24k - 3 \\
n = 3(8k - 1)
Only a factor of 3 comes out, not 8.
I am confused in one thing, if n/ (2^3, 3) gives a remainder of 21 which has factors of 7 and 3 then does it not imply that n must have 2^3 because it is not in remainder and it cancels out?
No, as 21 is the remainder and not the result or quotient.
If we take 21 as quotient instead then
n/(2^3 * 3) = 3 * 7
n = 2^3 * 3^2 * 7
this way n would have 2^3
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