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Manager
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If P, Q, and R are distinct positive digits and the product of the two
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18 Oct 2016, 09:11
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If P, Q, and R are distinct positive digits and the product of the twodigit 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
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18 Oct 2016, 09:36
sandeepnb wrote: If P, Q, and R are distinct positive digits and the product of the twodigit 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
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22 Nov 2016, 09:40
rohit8865Factoring 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
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22 Nov 2016, 10:21
championspud wrote: rohit8865Factoring out 221 is time consuming, would take more than 2 minutes..any fast method? championspudsee 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 1119... 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
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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
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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^\({2}\)+P2 = 0 => (P1)(P+2)=0 => P=1 or P = 2. Since P > 0 => P=1 => P+Q+R = 3+7+1 = 11




Re: If P, Q, and R are distinct positive digits and the product of the two
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24 Aug 2019, 19:57






