When x is divided by 3, the remainder is 1. When x is divided by 7, th

Question

When x is divided by 3, the remainder is 1. When x is divided by 7, the remainder is 2. How many positive integers less than 100 could be values for x?

A. 2
B. 4
C. 5
D. 6
E. 7

doomedcat · Accepted Answer

x=3a+1; = 1,4,7,10,13, 16 
x=7b+2;= 2,9, 16

combined we have x=lcm(3,7)+ 16 (common x) = 21p+16 , put p=0,1,2...

we have x = 16,37,58,79 (stop only as next numbers would be >=100)

so B 4#

frrcattack · Answer

Answer B: 4

First equation X=3Q+1 for some number Q
Second equation X=7N+2 for some number N

Using the negative remainders technique, I "added" 7 to N in the second equation to get the equation X=7T-5 where 7T=7N+7

I also "added" 3 twice to the first equation to get X=3Z-5 where 3Z=3Q+6 (add 3 twice to go from remainder -2 to -5)

Take X=7T-5 and X=3Z-5 and move the 5 to the other side for each equation.

Now X+5 must be some number that is divisible by both 7 and 3. So X could be 16, 37, 58, or 79... 4 choices so answer choice B.

baru · Answer

given X div by 3 - rem is 1 ==> x could be =  1,4,7,10,13,16,19,22,25,28 , 31,34,37,40,43,46,49,52,55,58 ,...pattern ( 1+30 ,4+30,7+30,... )
 X div by 7 - rem is 2 ==> x could be = 3,16,23,30,37,44,51,58,65,72,79,86,93(check - 65-30 =  35 ,72-30 =  42 ,79-30 =  49 )
So common numbers are {16,37,58,79} - option B

gracie · Answer

x=3q+1
x=7p+2
→7p-3q=-1
plugging in 0,1,2...for p,
least values for p and q are 2 and 5
so least value of x=16
let n=number of positive integer values for x<100
16+(n-1)(3*7)<100
→21n<105
n<5
n=4
B

Asianyc · Answer

3n=7m +1 
M= 2,5,8,11
So 4 numbers 7m+2 : 16,37,58,79

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taniad · Answer

we know that 7n+2=3a+1 (for some number n and a respectively)
hence, 3a-7n= 1

I quickly listed out the multiples of 7 and 3 upto 100 and compared it to see when the above held true, hence 4 instances:

3(5)-7(2)=1
3(9)-7(4)=1
3(16)-7(7)=1
3(33)-7(14)= 1

Aravind04 · Answer

Extending series 1 we get 1,4,10,13, 16 ,19,22,25,31,34, 37 -----97 
Extending series 2 we get 2,9, 16 ,23,30, 37 -----100

Series 1 has common number with series 2 for every 7th number
Series 2 has common number with series 1 for every 3rd number

Total digits in series 1 is (last number -1st number)/frequency ==>(97-1)/3==>32 digits
Total digits in series 2 is (last number -1st number)/frequency ==>(100-2)/7==>14 digits

32/7= quotient =4 or 14/3=quotient =4

hence there are 4 digits below 100

BrushMyQuant · Answer

Given that When x is divided by 3, the remainder is 1. When x is divided by 7, the remainder is 2. And we need to find How many positive integers less than 100 could be values for x

Theory: Dividend = Divisor*Quotient + Remainder

When x is divided by 3, the remainder is 1

x -> Dividend
3 -> Divisor
a -> Quotient (Assume)
1 -> Remainders
=> x = 3*a + 1 = 3a + 1

When x is divided by 7, the remainder is 2

x -> Dividend
7 -> Divisor
b -> Quotient (Assume)
2 -> Remainders
=> x = 7*b + 2 = 7b + 2

x = 3a + 1 = 7b + 2
=> a =  \frac{7b + 1}{3}

Only those values of b which will also give a as integer will give us the common values of x
 b = 2, 5, 8, 11, 14,...
But for b = 14, we will get x = 7b + 2 = 7*14 + 2 = 100 which is NOT less than 100

=> 4 values of x less than 100 are possible

So,  Answer will be B 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

rak08 · Answer

very time consuming. can someone suggest a shorter way
i listed all 3 7 tables for this

Krunaal · Answer

This is the fastest method, by making a common general form of remainder equation: https://gmatclub.com/forum/when-x-is-di ... l#p2186201

ravi1522 · Answer

Hello,
X=3I+1
X=7Y+2
so to find common term
3I+1=7Y+2
I= (7Y+1)/3 
so Y=2, I=5
first common term will be 16
X=21I+16
so I can take 0,1,2,3

so X can be 4 nos less than 100

Hence option B is correct
I hope it helps

When x is divided by 3, the remainder is 1. When x is divided by 7, th

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When x is divided by 3, the remainder is 1. When x is divided by 7, th

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