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# If r and s are non negative integers, p = 1.9*10^r and q = 1.9*10^s

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Manager
Joined: 07 Jun 2018
Posts: 51
Location: United States
If r and s are non negative integers, p = 1.9*10^r and q = 1.9*10^s  [#permalink]

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04 Nov 2018, 09:33
3
00:00

Difficulty:

55% (hard)

Question Stats:

55% (02:11) correct 45% (02:07) wrong based on 32 sessions

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If r and s are non negative integers, $$p=1.9*10^r$$ and $$q = 1.9*10^s$$, what is the units digit of 10(p+q)?

(1) r is divisible by 2

(2) s is not divisible by 2
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Joined: 07 Dec 2017
Posts: 1155
Re: If r and s are non negative integers, p = 1.9*10^r and q = 1.9*10^s  [#permalink]

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04 Nov 2018, 10:33
depending on the values of s and r, p and q each have a units digit of either 1 (if r or s are 0, making the value 1.9), 9 (if r/s are 1, making the value 19), or 0 (if r/s >1, for which the value will be 190, 1900...etc)
Clearly, both statements on their own are insufficient, since each only gives information on one of r and s, and not on the other.
Combined: from 1), p's units digit is 0 (r=2 ==> 19, r = 4 ==> 19,00, etc). from 2), q's units digit can either 1 (s=0), 9 (s=1), or 0 (s>2). specifically, if s=0, then P+q will have a decimal digit of 9 (x.9), and 10(p+q) will have a units digit of 9. Since we don't if this is the case or not, this is insufficient.
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Re: If r and s are non negative integers, p = 1.9*10^r and q = 1.9*10^s  [#permalink]

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04 Nov 2018, 10:33
Question : Unit digit of 10(p+q) = $$10(1.9*10^r+1.9*10^s)$$ = $$19(10^r+10^s)$$

From statement 1:

r is divisible by 2.
Then r could be 0,2,4,6...
If r is 0 then 10^0 = 1.
If r is 2 then 10^1 = 10.
Insufficient.

From statement 2:

s is not divisible by 2.
Then s could be 1,3,5...
Again r is unknown.
Insufficient.

Combining both:
Let r = 0. s = 1.
10^0 = 1 and 10^1 = 10. then 19(11) = last digit is 9.
If r = 2. and s= 3.
Then unit digit will be 0.
Combining is also insufficient.

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Re: If r and s are non negative integers, p = 1.9*10^r and q = 1.9*10^s   [#permalink] 04 Nov 2018, 10:33
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