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Re: The units digit of J vs The units digit of K [#permalink]
2
Popo wrote:
Can we not simply multiply the pattern of all numbers?

QA
We know that 2 has the 4 following patterns: 2,4,8,6 so (2^43) must have an ending of 8
3 has the following pattern 3,9,7,1 so 3^44 must be 1
5 has always the same pattern 5.
QB
7: 7,9,3,1 so 7^46 is 9
11 is like 5. The same pattern so always 1.
13: 3,9,7,1 so 7^48 so the ending is 1.

QA : (....8)(....1)(....5) = end with 0
QB: (....9)(....1)(....1) = end with 9

So B is greater


That works too.

Cheers,
Brent
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Re: The units digit of J vs The units digit of K [#permalink]
1
J=Units digit is "0"
K= Units digit which is greater than 0
So, answer is B
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Re: The units digit of J vs The units digit of K [#permalink]
J = 2^43 * 3^44 * 5^45 = (2*3*5)^43 * 3^1 * 5^2 = 30^43 * 3 * 25 --> 30 times anything will always end in "0"

K = 7^46 * 11^47 * 13^48 = (7*11*13)^46 * 11^1 * 13^2 = 1001^46 * 1859 --> since 1001 times itself repeating will always end in 1s, then multiplied by 1859 if will end in a "9"

K>J --> answer choice = B
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Re: The units digit of J vs The units digit of K [#permalink]
Hello from the GRE Prep Club BumpBot!

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Re: The units digit of J vs The units digit of K [#permalink]
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