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26 Apr 2016, 01:00
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A
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Difficulty:
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If N = 775 × 778 × 781, what is the remainder when N is divided by 14?
A. 6
B. 7
C. 8
D. 9
E. 10
A. 6
B. 7
C. 8
D. 9
E. 10
22 Aug 2018, 06:49
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20 seconds approach.
Ans A
Consider kudos if that helped.
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Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
Updated on: 26 Apr 2016, 17:03
Originally posted by InstantMBA on 26 Apr 2016, 04:55.
Last edited by InstantMBA on 26 Apr 2016, 17:03, edited 1 time in total.
Last edited by InstantMBA on 26 Apr 2016, 17:03, edited 1 time in total.
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Each of the three numbers may be written as \(14p+a\), where p is an integer and a is the remainder we’re looking for. They’d look like this:
\(775 = (14p + a)\)
\(778 = (14q + b)\)
\(781 = (14r + c)\)
Example #1:
Now, lets multiple two of them and identify the remainder of that product.
\(775 * 778\)
\(= (14p + a)(14q + b)\)
\(= 14^2pq + 14qa + 14pb + ab\)
\(= 14 (14pq + qa + pb) + ab\)
All but one term are divisible by 14. So the remainder of that product (775*778) divided by 14 is the remainder of the product of their remainders (ab) divided by 14 only. Here, the remainder of 775/14 is 5 and the remainder of 778/14 is 8. Their product (ab) = 5*8 = 40. The remainder of 40/14 is 12. So the remainder of 775*778/14 is the same, 12.
Example #2:
Similarly,
\(775 * 778 * 781\)
\(= (14p + a)(14q +b)(14r + c)\)
\(= 14 (14^2pqr + 14pqc + 14pbr + 14aqr + pbc + aqc + abr) + abc\)
Again all but one term are divisible by 14. In other words, it's the remainder of that product of the numbers/14 is the remainder of abc/14, only.
General rule:
The remainder of the product of three large numbers divided by \(x\) is the remainder of the product of their remainders divided by \(x\).
Answer to the OP:
The following are easy to determine:
The remainder of 775/14 is 5
The remainder of 778/14 is 8
The remainder of 781/14 is 11
Finally, the remainder of (775 * 778 * 781)/14 is simply the remainder of the product of their remainders divided by 14:
\(abc / 14\)
\(= (5 * 8 * 11)/14\)
\(= 440/14\)
\(= 31 + 6/14\)
The remainder is thus 6, which corresponds to A.
\(775 = (14p + a)\)
\(778 = (14q + b)\)
\(781 = (14r + c)\)
Example #1:
Now, lets multiple two of them and identify the remainder of that product.
\(775 * 778\)
\(= (14p + a)(14q + b)\)
\(= 14^2pq + 14qa + 14pb + ab\)
\(= 14 (14pq + qa + pb) + ab\)
All but one term are divisible by 14. So the remainder of that product (775*778) divided by 14 is the remainder of the product of their remainders (ab) divided by 14 only. Here, the remainder of 775/14 is 5 and the remainder of 778/14 is 8. Their product (ab) = 5*8 = 40. The remainder of 40/14 is 12. So the remainder of 775*778/14 is the same, 12.
Example #2:
Similarly,
\(775 * 778 * 781\)
\(= (14p + a)(14q +b)(14r + c)\)
\(= 14 (14^2pqr + 14pqc + 14pbr + 14aqr + pbc + aqc + abr) + abc\)
Again all but one term are divisible by 14. In other words, it's the remainder of that product of the numbers/14 is the remainder of abc/14, only.
General rule:
The remainder of the product of three large numbers divided by \(x\) is the remainder of the product of their remainders divided by \(x\).
Answer to the OP:
The following are easy to determine:
The remainder of 775/14 is 5
The remainder of 778/14 is 8
The remainder of 781/14 is 11
Finally, the remainder of (775 * 778 * 781)/14 is simply the remainder of the product of their remainders divided by 14:
\(abc / 14\)
\(= (5 * 8 * 11)/14\)
\(= 440/14\)
\(= 31 + 6/14\)
The remainder is thus 6, which corresponds to A.
