Aptitude - Numbers - Discussion

Hi guys,

Just remember these formula's and verified it yourself.

a^n-b^n is divisible by a-b for all n.
a^n-b^n is divisible by a+b if n is Even.
a^n+b^n is divisible by a+b if n is Odd.

In our questions 49^15-1 is divisible by 48 as per formula but option is not available. So we have to further drill down to get the answer. Further simplification it became 7^30-1.

From the above formula answer would be 6, 8 and 6 is not available in the options. So answer is 8.

From divisibilty tricks,
1.-> (a^n-b^n) is always divisible by (a-b).
2.-> (a^n-b^n) is divisible by (a+b) when 'n' is even.

In the question,
(49^15-1), it could also be written as (49^15-1^15) as,(1^n=1) but '49' is the square of '7'. By simplifying, we have,
(7^2)^15-(1^)^15 as, 7^2=49 and 1^n=1.
= 7^30-1^30 (here n=30 is an even number).

So, by using formula 2. We have,
= 7+1,
= 8 (answer).

9 power odd number results in 9 at units place.
9 power even number results in 1 at units place.

In 49 or 9 power 15.... 15 is odd therefore 9 will come at units place.
9-1 = 8 which is divisible by option[A] i.e., 8

@Prasanna.

48 is not possible because rules tell that if x power n subtract a power n then result is x power n please a power. So I can say that may be the 50 answer is also possible.

(x^n-1) is divisible by x-1 when n is odd
Given
(49^15-1) is divisible by 49-1 i.e 48
48 is completely divisible by 8.(Here 15 is odd right)

Can we do it in this way?

49^15 -1
with the help of cyclicity table the unit digit will be 9.

Unit digit for 49^15 = 9.

Now 9 -1 (given in question) = 8 answer.

Simpliy, find the unit digit of 49^15, which is 49^15 = 9.
Then,(9-1) = 8.
Thank you very much!

X^n-y^n=x-y.

So,
49^15-1^15=49-1=48.
48÷8....48 is divisible by 8 so, 8 is answer.

Can anyone please explain how to find 7 power 6-4 power 6 is divisible by?

a^n-b^n=a-b i.e 49-1=48 which is divisible by 8.

So the answer is 8.