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# If x, y, and z are single-digit integers and 100(x) + 1,000(y) + 10(z)

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If x, y, and z are single-digit integers and 100(x) + 1,000(y) + 10(z)  [#permalink]

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30 Nov 2018, 06:04
00:00

Difficulty:

5% (low)

Question Stats:

94% (00:41) correct 6% (00:40) wrong based on 112 sessions

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If x, y, and z are single-digit integers and 100(x) + 1,000(y) + 10(z) = N, what is the units' digit of the number N?

(A) 0
(B) 1
(C) x
(D) y
(E) z

Project PS Butler : Question #49

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Re: If x, y, and z are single-digit integers and 100(x) + 1,000(y) + 10(z)  [#permalink]

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30 Nov 2018, 08:09
1
10*z = z0, 100*x = x00, 1000*y = y000

Sum will be y000 + x00 + z0 = yxz0

So units digit will be 0.

Option A
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Re: If x, y, and z are single-digit integers and 100(x) + 1,000(y) + 10(z)  [#permalink]

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30 Nov 2018, 15:16
Since we know that tens' and units' digits of N can be found in the product of $$10*z$$ and that z is a single-digit integer, we can conclude that every allowable integer (i.e. from 0 to 9) multiplied by 10 will end with 0 (e.g. $$1*10=10$$, $$6*10=60$$, $$9*10=90$$). Hence, the correct answer is A.

We can also plug in possible numbers and check the result.

Let x=1, y=2, z=3:

$$N = 100*1+1000*2+10*3 = 100+2000+30 = 2130$$

It's clear that N 's last digit will always be 0.
Re: If x, y, and z are single-digit integers and 100(x) + 1,000(y) + 10(z)   [#permalink] 30 Nov 2018, 15:16
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