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# Find the units digit of (556)^17n+ (339)^(5m+15n), where m and n are

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Find the units digit of (556)^17n+ (339)^(5m+15n), where m and n are  [#permalink]

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Updated on: 24 Nov 2016, 12:14
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Question Stats:

56% (02:14) correct 44% (02:34) wrong based on 89 sessions

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Find the units digit of (556)^17n + (339)^(5m+15n) , where m and n are positive integers

(1) 4m+12n = 360
(2) n is the smallest 2-digit positive integer divisible by 5

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Originally posted by stonecold on 23 Nov 2016, 08:37.
Last edited by stonecold on 24 Nov 2016, 12:14, edited 7 times in total.
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Re: Find the units digit of (556)^17n+ (339)^(5m+15n), where m and n are  [#permalink]

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23 Nov 2016, 09:05
1
556^17n will always have 6 in unit's digit.
5m+15n=5(m+3n)
339^(5m+15n) can have units digit 9 or 1 depending on whether (5m+15n) is odd or even.
S1 - 4m+12n = 360
4(m+3n)=360
m+3n = 90 i.e. even
So 5m+15m is even, hence units digit can be found.

S2 - n is the smallest 2-digit positive integer divisible by 5 i.e. 10.
but m can be any value. so, units digit cannot be found

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Re: Find the units digit of (556)^17n+ (339)^(5m+15n), where m and n are  [#permalink]

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24 Nov 2016, 05:14
1
stonecold wrote:
Find the units digit of (556)^17n + (339)^(5m+15n), where m and n are positive integers

(1) 4m+12n = 360
(2) n is the smallest 2-digit positive integer divisible by 5

556 has the units digit of 6 so it will always end in 6. We don't need to worry about the exponent here.

339 has the units digit of 9 so its cyclicity is 9, 1, 9, 1, 9, 1, 9, 1... etc.
If the exponent is odd, 339 will end with 9. If the exponent is even, 339 will end with 1.

We need to find whether (5m+15n) is odd or even.

(1) 4m+12n = 360
m + 3n = 90
There are only two ways in which we can get an even sum.
Odd + Odd = Even or
Even + Even = Even
So m and n are either both odd or both even.
Hence 5m +15n must be even only. So 339^{5m + 15n} ends with a 1.

Hence, (556)^17n + (339)^(5m+15n) will have units digit of 6+1 = 7.
Sufficient

(2) n is the smallest 2-digit positive integer divisible by 5
n must be 10 - even
But we don't know about m. If it is odd, (5m+15n) will be odd. If it is even, (5m+15n) will be even.
Not sufficient.

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Re: Find the units digit of (556)^17n+ (339)^(5m+15n), where m and n are  [#permalink]

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24 Nov 2016, 11:51
stonecold wrote:
Find the units digit of (556)^17n + (339)^(5m+15n), where m and n are positive integers

(1) 4m+12n = 360
(2) n is the smallest 2-digit positive integer divisible by 5

$$(556)^{17n} + (339)^{5m+15n}$$

$$(556)^{17n} + (339)^{5(m+3n)}$$

$$(556)^{17n}$$ will always have the units digit as 6 $$( As \ 6^{ Any \ Number })$$ is 6

FROM STATEMENT - I ( SUFFICIENT )

$$4m+12n = 360$$

Or, $$m + 3n = 90$$

Now, go for units digit of $$(556)^{17n} + (339)^{5(m+3n)}$$ = $$6 + 9^{5*90}$$

$$9^{ Even }$$ = Units digit 1
$$9^{ Odd }$$ = Units digit 9

So, $$9^{450}$$ will have units digit as 1

So, The units digit of $$(556)^{17n} + (339)^{5m+15n}$$ will be $$6 + 1 =7$$

FROM STATEMENT - II ( INSUFFICIENT )

We know units digit of $$(556)^{17n}$$ will be 6

Least 2 digit positive integer divisible by 5 is 10

Now, $$(339)^{5(m+3n)} = (339)^{5(m+3*10)} = (339)^{5(m+30)}$$ , but we do not know the value of m so we can not find the units digit of of $$9^{5(m+3*10)}$$

Thus statement 2 is insufficient....

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked, answer will be (A)

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Re: Find the units digit of (556)^17n+ (339)^(5m+15n), where m and n are  [#permalink]

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28 Feb 2018, 20:56
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Re: Find the units digit of (556)^17n+ (339)^(5m+15n), where m and n are   [#permalink] 28 Feb 2018, 20:56
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