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23 Jun 2020, 07:39
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For which of the following values of n will the remainder of 351^n and 352^n be the same when divided by 7?
I. 3
II. 4
III. 6
A. I only
B. II only
C. I and III only
D. II and III only
E. I, II, and III
Are You Up For the Challenge: 700 Level Questions
I. 3
II. 4
III. 6
A. I only
B. II only
C. I and III only
D. II and III only
E. I, II, and III
Are You Up For the Challenge: 700 Level Questions
yashikaaggarwal
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23 Jun 2020, 08:07
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To find : for what value of n the remainder of 351^n/7 And 352^n/7 is N
Solution: (350+1)^n/7 = 1^n/7
=> (350+2)^n/7 = 2^n/7
I. 3
=> 1^3 mod 7 = 1 as remainder
=> 2^3 mod 7 = 8 mod 7 = 1 as remainder. (Matching)
II. 4
=> 1^4 mod 7 = 1 as remainder
=> 2^4 mod 7 = 16 mod 7 = 2 as remainder. (Not Matching)
III. 6
=> 1^6 mod 7 = 1 as remainder
=> 2^6 mod 7 = 64 mod 7 = 1 as remainder. (Matching)
Only statement 1&3 are matching.
IMO C
Posted from my mobile device
Solution: (350+1)^n/7 = 1^n/7
=> (350+2)^n/7 = 2^n/7
I. 3
=> 1^3 mod 7 = 1 as remainder
=> 2^3 mod 7 = 8 mod 7 = 1 as remainder. (Matching)
II. 4
=> 1^4 mod 7 = 1 as remainder
=> 2^4 mod 7 = 16 mod 7 = 2 as remainder. (Not Matching)
III. 6
=> 1^6 mod 7 = 1 as remainder
=> 2^6 mod 7 = 64 mod 7 = 1 as remainder. (Matching)
Only statement 1&3 are matching.
IMO C
Posted from my mobile device
chrtpmdr
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30 Jul 2020, 03:09
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Note that there is an inheritent pattern. Do solve such questions untimed, I usually look at the same principle tested with easier numbers.
Lets pick a number that gives a remainder of 2 when being divided by 7 -> 2
2 : 7-> R:2 let's replicate the pattern by raising the dividend by 2
4 : 7 -> R:4 first hint: the remainder was raised by 2 as well
8 : 7 -> R:1 if we would again raise the initial remainder by 3 (2^3) we would receive 8 as the remainder. As we are dividing by 7 we know that the remainder can never be greater or equal than the divisor, thus 8/7 -> R:1
16 : 7 -> R:2 at this point we should realize the patter 2-4-1
Same is applicable with 1 just that the remainder will always remain at one. We are thus looking for N such that it will result in a remainder of 1 for both expression. Remember that every 3rd remainder in the pattern was one. Thus we can safely assume that if n = 3 -> R:1 and if N = 6 -> R:1
Lets pick a number that gives a remainder of 2 when being divided by 7 -> 2
2 : 7-> R:2 let's replicate the pattern by raising the dividend by 2
4 : 7 -> R:4 first hint: the remainder was raised by 2 as well
8 : 7 -> R:1 if we would again raise the initial remainder by 3 (2^3) we would receive 8 as the remainder. As we are dividing by 7 we know that the remainder can never be greater or equal than the divisor, thus 8/7 -> R:1
16 : 7 -> R:2 at this point we should realize the patter 2-4-1
Same is applicable with 1 just that the remainder will always remain at one. We are thus looking for N such that it will result in a remainder of 1 for both expression. Remember that every 3rd remainder in the pattern was one. Thus we can safely assume that if n = 3 -> R:1 and if N = 6 -> R:1
