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I have collected these problems on remainder. This type of problem is frequently asked in DS.
Answers are also given. Please dont mind any typo error.
1.If r is the remainder when the positive integer n is divided by 7, what is the value of r
1. when n is divided by 21, the remainder is an odd number
2. when n is divided by 28, the remainder is 3
The possible reminders can be 1,2,3,4,5 and 6. We have the pinpoint the exact remainder from this 6 numbers.
St 1: when n is divided by 21 ( 7 and 3) the remainder is an odd number.
But it cannot be 7, 3 or 9 . Hence the possibilities are : 1 and 5.
Hence there can be two remainders ,1 and 5, when divided by 7.
NOT SUFFICIENT
St 2: when n is divided by 28 the remainder is 3.
As 7 is a factor of 28, the remainder when divided by 7 will be 3
SUFFICIENT
2 If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?
(1) n = 2
(2) m = 1
The Concept tested here is cycles of powers of 3.
The cycles of powers of 3 are : 3,9,7,1
St I) n = 2. This makes 3^(4*2 +2 + m) = 3^(10+m). we do not know m and hence cannot figure out the unit digit.
St II) m=1 . This makes 3^(4*n +2 + 1).
4n can be 4,8,12,16...
3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7. SUFF
Hence B
3.If p is a positive odd integer, what is the remainder when p is divided by 4 ?
(1) When p is divided by 8, the remainder is 5.
(2) p is the sum of the squares of two positive integers.
st1. take multiples of 8....divide them by 4...remainder =1 in each case...
st2.
p is odd ,since p is square of 2 integers...one will be even and other
odd....now when we divide any even square by 4 v ll gt 0 remainder..and
when divide odd square vll get 1 as remainder......so intoatal
remainder=1
Ans : D
4.If p and n are positive integers and p > n, what is the remainder when p^2 - n^2 is divided by 15 ?
(1) The remainder when p + n is divided by 5 is 1.
(2) The remainder when p - n is divided by 3 is 1.
Ans: E
st1) p+n=6,11,16....insuff.
st2) p-n=4,7,10....insuff...
multiply these two to get p^2-n^2.....multiplying any ttwo values from the above results in different remainder......
also can be done thru equation....p+n=5a+1..and so on
5.What is the remainder when the positive integer x is divided by 3 ?
(1) When x is divided by 6, the remainder is 2.
(2) When x is divided by 15, the remainder is 2.
Easy one , answer D
st 1...multiple of 6 will also be multiple of 3 so remainder wil be same as 2.
st2)
multiple of 15 will also be multiple of 3....so the no.that gives
remaindr 2 when divided by 15 also gives 2 as the remainder when
divided by 3...
6.What is the remainder when the positive integer n is divided by 6 ?
(1) n is a multiple of 5.
(2) n is a multiple of 12.
Easy one. Answer B
st 1) multiples of 5=5,10,15....all gives differnt remainders with 6
st2)n is divided by 12...so it will be divided by 6...remainder=0
7If x, y, and z are positive integers, what is the remainder when 100x + 10y + z is divided by 7 ?
(1) y = 6
(2) z = 3
We need to know all the variables. We cannot get that from both the statements. Hence the answer is E.
8.If n is a positive integer and r is the remainder when 4 + 7n is divided by 3, what is the value of r ?
(1) n + 1 is divisible by 3.
(2) n > 20
Answer A
st1... n+1 divisible by 3..so n=2,5,8,11......
this gives 4+7n=18,39,60....remainder 0 in each case......
st2) insufficient ....n can have any value
9.If n is a positive integer and r is the remainder when (n - 1)(n + 1) is divided by 24, what is the value of r ?
(1) n is not divisible by 2.
(2) n is not divisible by 3.
ST
1- if n is not divisible by 2, then n is odd, so both (n - 1) and (n +
1) are even. moreover, since every other even number is a multiple of
4, one of those two factors is a multiple of 4. so the product (n -
1)(n + 1) contains one multiple of 2 and one multiple of 4, so it
contains at least 2 x 2 x 2 = three 2's in its prime factorization.
But
this is not sufficient, because it can be (n-1)*(n+1) can be 2*4 where
remainder is 8. it can be 4*6 where the remainder is 0.
ST 2-
if n is not divisible by 3, then exactly one of (n - 1) and (n + 1) is
divisible by 3, because every third integer is divisible by 3.
therefore, the product (n - 1)(n + 1) contains a 3 in its prime
factorization.
Just like st 1 this is not sufficient
the overall prime factorization of (n - 1)(n + 1) contains three 2's and a 3.
therefore, it is a multiple of 24.
sufficient
Answer C
