If a and b are positive integers, is the sum a + b divisible by 4?

Question

I. When the sum 23^a+25^b is divided by 10, the remainder is 8
II. When 22^b is divided by 10, the remainder is 8

m1033512 · Answer

statement 1

we get remainder 8 when sum of last digits of both the terms is 8

so a can take values 1,5,..

where we always get last digit as 3 
and b can take any value 1,2,3 and so on becuse here we will always get 5 as last digit.

so for a =1 and b=1 , it is not divisible

but for a= 5 and b=3,

it is divisible,

Not sufficient

statement 2

we get values of b as 3, 7 , 11 and so on
but it does not say anything about a 
so insufficient

combined together

we get a = 1,5 ...
b= 3,7,.....

so here a+b is divisible by 4

Answer C

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avinashnerella · Answer

When the sum 23^a + 25^b is divided by 10, the remainder is 8, we can say that unit digit should be 8.
For all values of b, unit’s of 25^b is 5 only.
Then 23^a’s unit’s should be 3. It is possible for a = 1 or 5 or 9 etc. 
Insufficient.
Statement (2):
When 22^b is divided by 10, the remainder is 8, we can say that unit’s digits should be 8.
Unit’s digit is 8 for b = 3 or 7 or 11 etc.
Insufficient.
Together:
a = 1 or 5 or 9 etc and b = 3 or 7 or 11 etc.
a divided by 4 , remainder is 1 , b divided 4 remainder is 3. Then (a+b) divided by 4 remainder is Zero.
Sufficient.
Answer: C.

chetan2u · Answer

So we require to know the value of a+b or the last two digits as divisibility by 4 depends on last two digits.

(1)  23^a+25^b  divided by 10 gives remainder 8.
This means that the last digits of a and b add up to 8. 
25^b will leave a remainder 5, whatever be the value of b. 
To get the sum as 8, we should get the last digit of 23^a as 3. The cycilicty of 3 is 3,9,7,1,3,9,7,1..., so a can be 1,5,9.. or a has to be of type 4K+1.
If b is 3,7... or of type 4x-1, answer is yes, otherwise no. 
Insuff

(2) 22^b divided by 10 gives a remainder 8.
Cyclicity of 2 is 2,4,8,6,2,4,8,6.... But we are looking for last digits as 8, so b can be 3,7.....or of type 4b-1.
Nothing known about a. 
Insuff

Combined
a is of type 4K+1 and b is of type 4b-1, so a+b=4K+1+4b-1=4(k+b), which is divisible by 4.
Sufficient

C

Archit3110 · Answer

a+b sum divisible by 4
#1
When the sum 23^a+25^b is divided by 10, the remainder is 8
possible when a is odd integer values possible 1,5,9,13
and b is odd integer value ; 1,3,5,7,9..
we get yes & no a+b ; 1+1 ; no 5+3 ; yes
insufficient
#2
When 22^b is divided by 10, the remainder is 8
possible with values of b being 3,7,11,15 
value of a not know insufficient
from 1 &2
when b is 3 a would be 5 i.e sum of both is divisible by 4
sufficient
option C

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If a and b are positive integers, is the sum a + b divisible by 4?

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If a and b are positive integers, is the sum a + b divisible by 4?

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