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# Around the World in 80 Questions (Day 3): If x and y are consecutive

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Re: Around the World in 80 Questions (Day 3): If x and y are consecutive [#permalink]
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If x and y are consecutive positive two-digit integers, what is the value of x + y?

(1) When x^25 is divided by y the remainder is 11
(2) When x^36 is divided by y the remainder is 1

case I:
x^25 = y*a + 11
thus y = 12, 13....
now if y = 12, x = 11 or 13
x = 11. thus 11^25/12 remainder is -1 = 11
x = 13. thus 13^25/12 remainder is 1

now if y = 13, x = 12 or 14
x = 12. thus 12^25/13 remainder is -1 = 12
x = 14. thus 14^25/13 remainder is 1

case II:
x^36 = b* y + 1
thus if y = 12, x = 11 or 13
x = 11^36/12, remainder = 1
x = 13^36/12, remainder = 1

thus if y = 11, x = 10 or 12
x = 10^36/11, remainder = 1
x = 12^36/11, remainder = 1

thus x= 11, y = 12

thus A
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Re: Around the World in 80 Questions (Day 3): If x and y are consecutive [#permalink]
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Kudos
here there are 2 possiblities of x and y ( they are consecutive int )
if x>y then x raised to anything divided by y will give us 1 as remainder
and if x<y then x raised to anything divided by y will give us -1 as remainder
I found this link to be helpful :- https://prepinsta.com/how-to-find-remai ... s-quickly/

1. when x^25 is divided by y then remainder is 11 now this case we have 2nd condition
hence -1 will translate to y = 12 and x should be less than y hence x= 11
suff

2. when x^36 is divided by y then remainder is 1.
here we can have any 2 consecutive int x and y such that x>y we will get remainder 1
insuff

Ans A
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Re: Around the World in 80 Questions (Day 3): If x and y are consecutive [#permalink]
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If x and y are consecutive positive two-digit integers, what is the value of x + y?

(1) When x^25 is divided by y the remainder is 11
(2) When x^36 is divided by y the remainder is 1

Solution:

Given x and y are consecutive integers.

(1) When x^25 is divided by y the remainder is 11
Case I: x>y
In this case, the remainder should be 1 and even 1^25 = 1. But it is given that the remainder is 11. Hence, this case is not possible.

Case II: x<y
In this case, remainder is (-1) and (-1)^25 is (-1). But because there doesn't exist any concept of negative remainders hence this remainder is subtracted from the divisor (y in this case) to arrive at the final remainder which is given as 11. So y-1=11 or y=12. Also, x<y and both are consecutive so x=11. Therefore, x+y=23 (can be calculated).

SUFFICIENT.

(2) When x^36 is divided by y the remainder is 1
Case I: x>y
In this case, the remainder should be 1 and even 1^36 = 1. It leaves us infinite possibilities for possible pairs of x and y. x+y cannot be calculated. INSUFFICIENT.

Case II: x<y
In this case, remainder is (-1) and (-1)^36 is 1. This again leaves us infinite possibilities for possible pairs of x and y. x+y cannot be calculated. INSUFFICIENT.

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Re: Around the World in 80 Questions (Day 3): If x and y are consecutive [#permalink]
1
Kudos
If x and y are consecutive positive two-digit integers, what is the value of x + y?

(1) When x^25 is divided by y the remainder is 11

Only option is x=11, and y =12

11^25 mod 12
= (-1)^25 mod 12 (Note :- 11 mod 12 = 11 (or, -1 in terms of -ve remainder))
= (-1) mod 12
= -1
= -1+12
= 11

So, x+y =11 +12 =23

Hence, SUFFICIENT

(2) When x^36 is divided by y the remainder is 1

11^36 mod 12 (x=11, y =12)
= (-1)^36 mod 12 (Note :- 11 mod 12 = 11 (or, -1 in terms of -ve remainder))
= 1 mod 12
= 1
So, x+y =11 +12 =23......(1)

Also, if
12^36 mod 13 (x=12, y =13)
= (-1)^36 mod 13 (Note :- 12 mod 13 = 12 (or, -1 in terms of -ve remainder))
= 1 mod 12
= 1
So, x+y =12 +13=25....(2)

Since, two values possible (as above) for x+y

Hence, INSUFFICIENT

(A) is the CORRECT answer
Re: Around the World in 80 Questions (Day 3): If x and y are consecutive [#permalink]
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