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What should be the OA for this one?

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by vaibhavob » Sun Jan 23, 2011 3:54 am
If X and Y are positive integers and n = 5^X and 7^(y+15), what is the units digit of n?
1.y = 2x - 15
2.y^2 - 6y + 5 = 0

The OA is B.

Can someone elaborate on how to solve this one? Seems like a simple problem of units digit using cyclicity?

Hope the question makes sense now.
Last edited by vaibhavob on Sun Jan 23, 2011 4:15 am, edited 1 time in total.
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by maihuna » Sun Jan 23, 2011 4:05 am
Q is not clear, what do u mean by n = 5^X and 7^*(y+15)?

ANy power of 5 ends in 5 so no value of x required

y^2 - 6y + 5 = 0 =>(y-1)(y-5)=0=>y=1 o r5

7^16 ends in (7^4)^4 = 1^4 = 1 (7,9,3,1 is cyclicitty of power of 7) in 1.
7^20 ends in (7^4)^5 = 1^5 = 1

so B is fine even though it gives 2 values as both ends in 1
Charged up again to beat the beast :)
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by Anurag@Gurome » Sun Jan 23, 2011 4:05 am
If X and Y are positive integers and n = 5^X and 7^*(y+15), what is the units digit of n?
The red part doesn't makes sense.
Can you check that part and edit accordingly?
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by vaibhavob » Sun Jan 23, 2011 4:19 am
maihuna wrote:Q is not clear, what do u mean by n = 5^X and 7^*(y+15)?

ANy power of 5 ends in 5 so no value of x required

y^2 - 6y + 5 = 0 =>(y-1)(y-5)=0=>y=1 o r5

7^16 ends in (7^4)^4 = 1^4 = 1 (7,9,3,1 is cyclicitty of power of 7) in 1.
7^20 ends in (7^4)^5 = 1^5 = 1

so B is fine even though it gives 2 values as both ends in 1
Thanks...you understood the question correctly.. i figured out where i was making the mistake.
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