How many positive integers less than 500 have a remainder of 1 when

gmatophobia can you please explain the concept after finding the first common term?

hope this helps: the LCM was taken and added with the lowest value. Then the given condition N<500 was substituted with the new arrived equation

\(N=7a+1=3b+2\\
N+13=7()=3()\\
N=LCM(3,7)-13=21z-13.\\
z=1,2....24\)

for \(0<N<500\)

Bunuel, do you have more such questions for practice?

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Hope it helps.

How many positive integers less than 500 have a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?

A. 20

B. 21

C. 24

D. 26

E. 72

When 7 divides 500 it gives 71 integers as quotient 497 being the last divisible number.
Hence there are 71 integers that leave remainder 1 498(497+1) being the last one.

When 3 divides 500 it gives 166 integers as quotient 498 being the last divisible number.
Hence there are 166 integers that leave remainder 1 499(498+2) being the last one.
But this is not helpful since we are looking for a common integer that leaves 1 and 2 remainder when divided by 7 and 3 individually.

So, only 71 integers are there among which common ones would leave remainders 1 and 2. These common can be found by dividing 71 by 3 which gives us 23.xx
Since we are looking for integer decimal is not possible hence we round up it to 24. Answer C.

Estimation also helps but that depends on the choices available. Had 23 been there among the choices it would have been difficult. But that is unlikely in these tests.

Answer C.

­How many positive integers less than 500 have a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?

When faced with a GMAT Quant question you're not sure how to answer, keep in mind that GMAT Quant questions are often designed to be answerable in multiple ways. So, it can often work to just try anything that might get you to the answer.

In this case, one approach that we can use is to look for a pattern starting with the smallest integer we can find such that it has a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3.

That integer is 8.

Then, we can continue with integers 1 greater than a multiple of 7 and look for a pattern to the ones that are 2 greater than a multiple of 3.

Doing so, we see the following:

8, 15, 22, 29, 36, 43, 50, 57, 64, 71

The integers in green fit our constraints. So, apparently, one in three integers that's 1 more than a multiple of 7 is two more than a multiple of 3.

There are 500 - 8 = 492 integers between 8 and 500.

492/7 = 70.X of those integers are one more than a multiple of 7.

70/3 = 23.X

23 + 1 (for 8) = 24 total integers that fit our constraints.

A. 20
B. 21
C. 24
D. 26
E. 72

Correct answer: C­