fast approach?

This has to do with the cyclicity of units digit for 7


Units digit of 7 cycle as 7,9,3,1,7,9,3,1

7^(4n+3)
n=0 then its 7^3
n=1 then 7^7
n=2 then 7^11

Always have 3 in the units digit

6^n will have always have 6 in the units digit (6,36,216 etc,,)


Units digit 3 from 7^(4n+3) form exponent * units digit of 6 from 6^n will result in a number with units digit 8

This divide by 10 will always leave a remainder 8

Hope I dint miss something here

E)

cramya wrote:This has to do with the cyclicity of units digit for 7


Units digit of 7 cycle as 7,9,3,1,7,9,3,1

7^(4n+3)
n=0 then its 7^3
n=1 then 7^7
n=2 then 7^11

Always have 3 in the units digit

6^n will have always have 6 in the units digit (6,36,216 etc,,)


Units digit 3 from 7^(4n+3) form exponent * units digit of 6 from 6^n will result in a number with units digit 8

This divide by 10 will always leave a remainder 8

Hope I dint miss something here

E)

Very nice, very nice!

bacali wrote:Can someone show me a fast way to do this. I actually multiplied it out, yes 7^7 * 6. I got it right but took me a while as you might imagine.

If n is a positive integer, what is the remainder
When (7^(4n+3))(6^n) is divided by 10?


A. 1
B. 2
C. 4
D. 6
E. 8



OA: E

I am not sure if i will be useful .. But iam using CRT

7^4n+3*6^n

put n= 2
7^11*6^2
7^11/10 ~ -7
6^2~ -4

thus its 28/10 ~ 8

hence E

sudhir3127 wrote:

bacali wrote:Can someone show me a fast way to do this. I actually multiplied it out, yes 7^7 * 6. I got it right but took me a while as you might imagine.

If n is a positive integer, what is the remainder
When (7^(4n+3))(6^n) is divided by 10?


A. 1
B. 2
C. 4
D. 6
E. 8



OA: E

I am not sure if i will be useful .. But iam using CRT

7^4n+3*6^n

put n= 2
7^11*6^2
7^11/10 ~ -7
6^2~ -4

thus its 28/10 ~ 8

hence E

Very helpful. Thanks.