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If P, Q, and R are distinct positive digits and the product of the two

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If P, Q, and R are distinct positive digits and the product of the two  [#permalink]

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New post 18 Oct 2016, 09:11
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If P, Q, and R are distinct positive digits and the product of the two-digit integers PQ and PR is 221, what is the sum of the digits P, Q, and R?

A. 5
B. 11
C. 13
D. 21
E. 23
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If P, Q, and R are distinct positive digits and the product of the two  [#permalink]

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New post 18 Oct 2016, 09:36
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sandeepnb wrote:
If P, Q, and R are distinct positive digits and the product of the two-digit integers PQ and PR is 221, what is the sum of the digits P, Q, and R?

A. 5
B. 11
C. 13
D. 21
E. 23


Factor out 221
221=13*17

Thus PQ=13 & PR = 17 or vice versa

Thus P=1, Q=3 & R=7
Sum=1+3+7=11

Ans B
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Re: If P, Q, and R are distinct positive digits and the product of the two  [#permalink]

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New post 22 Nov 2016, 09:40
rohit8865

Factoring out 221 is time consuming, would take more than 2 minutes..any fast method?
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Re: If P, Q, and R are distinct positive digits and the product of the two  [#permalink]

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New post 22 Nov 2016, 10:21
1
championspud wrote:
rohit8865

Factoring out 221 is time consuming, would take more than 2 minutes..any fast method?

championspud
see its given PQ and PR are two digit integers....

notice if P=2 then obvoiusly 2Q and 2R were greater than 400...
thus u can quickly chek numbers between 11-19...

now as 1 is last digit of 221 thus u can easily figure out further calculations...

hope it helps...

Rohit
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Re: If P, Q, and R are distinct positive digits and the product of the two  [#permalink]

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New post 22 Nov 2016, 15:17
221 = 17*13

Therefore, P = 1; Q= 7 and R = 3 => P+Q+R = 11
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Re: If P, Q, and R are distinct positive digits and the product of the two  [#permalink]

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New post 24 Aug 2019, 19:57
P ≠ Q ≠ R
P>0 ; Q>0; R>0
PQ * QR = 221 => Q*R must have unit digit = 1
Since Q ≠ R => Either Q=3 and R=7 or Q=7 and R=3

PQ*PR = 221
=> (10P+Q)(10P+R)=221
=> 100\(P^{2}\) + 10P(Q+R) + QR = 221
=> 100\(P^{2}\) + 10P*10 + 21 = 221
=> P^\({2}\)+P-2 = 0
=> (P-1)(P+2)=0
=> P=1 or P = -2. Since P > 0 => P=1
=> P+Q+R = 3+7+1 = 11
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Re: If P, Q, and R are distinct positive digits and the product of the two   [#permalink] 24 Aug 2019, 19:57
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