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07 Apr 2015, 06:12
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What are the last two digits of \((301*402*503*604*646*547*448*349)^2\)
(A) 96
(B) 76
(C) 56
(D) 36
(E) 16
Kudos for a correct solution.
(A) 96
(B) 76
(C) 56
(D) 36
(E) 16
Kudos for a correct solution.
13 Apr 2015, 07:29
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Bunuel
What are the last two digits of (301*402*503*604*646*547*448*349)^2
(A) 96
(B) 76
(C) 56
(D) 36
(E) 16
Kudos for a correct solution.
(A) 96
(B) 76
(C) 56
(D) 36
(E) 16
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:
We need to find the last two digits of the product. It means we need to find the remainder when we divide the product by 100.
Find remainder of (301*402*503*604*646*547*448*349)^2/100
Note that 301 = 300 + 1 which gives us a small remainder 1 to work with but 349 = 300 + 49, a large remainder with which calculations will become cumbersome. But note that 349 is close to 350. All the numbers in the product are quite close to a multiple of 50, if not to a multiple of 100.
We need to find the remainder of:
(301*402*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/100
This implies we need to find the remainder of:
(301*201*503*604*646*547*448*349)*(301*402*503*604*646*547*448*349)/50
We cancel off a 2 (of 402) from the numerator with a 2 of the denominator to make the divisor 50. We will multiply the remainder we obtain by 2 back at the end.
We need the remainder of:
(300 + 1)*(200+1)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)* (300+ 1)*(400+2)*(500+3)*(600+4)*(650-4)*(550-3)*(450-2)*(350-1)/50
Note that in this product, all terms obtained will have a multiple of 50 except the last term obtained by multiplying the remainders together:
We need:
the remainder of 1*1*3*4*(-4)*(-3)*(-2)*(-1)*1*2*3*4*(-4)*(-3)*(-2)*(-1) / 50
the remainder of (12)*(24)*(24)*(24) / 50
Notice now that the remainders are far too great but they are close to 25. So let’s bring the divisor down to 25 by canceling off another 2.
We need:
the remainder of (6)*(24)*(24)*(24) / 25
the remainder of (6)*(25-1)*(25-1)*(25-1) / 25
Again, the only product we need to worry about is the last one obtained by multiplying the remainders together:
the remainder of 6*(-1)*(-1)*(-1) / 25
The remainder is -6 which is negative. To get the positive remainder, 25 – 6 = 19. But remember that we had divided the divisor twice by 2 so to get the actual remainder, we must multiply the remainder obtained back by 4: the actual remainder is 4*19 = 76
Answer (B)
07 Apr 2015, 07:17
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What are the last two digits of (301*402*503*604*646*547*448*349)^2
One thing to note before I start: if you got 2 numbers with unknown digits except for last 2, you can then determine the last 2 digits of their product. For example, xxxxxx36 * xxxx47 = xxxxxxxx92 regardless of "x" where x - digits.
Lets change our number a little - as in, cut away useless information and leave only tenth and unit digits
\((301*402*503*604*646*547*448*349)^2\) -> \((1*2*3*4*46*47*48*49)^2\) -> \((24*46*47*48*49)^2\) -> \((4*47*48*49)^2\) -> \((88*48*49)^2\) -> \((24*49)^2\) -> \(76^2\) -> \(76\)- our last 2 digits
B that is
One thing to note before I start: if you got 2 numbers with unknown digits except for last 2, you can then determine the last 2 digits of their product. For example, xxxxxx36 * xxxx47 = xxxxxxxx92 regardless of "x" where x - digits.
Lets change our number a little - as in, cut away useless information and leave only tenth and unit digits
\((301*402*503*604*646*547*448*349)^2\) -> \((1*2*3*4*46*47*48*49)^2\) -> \((24*46*47*48*49)^2\) -> \((4*47*48*49)^2\) -> \((88*48*49)^2\) -> \((24*49)^2\) -> \(76^2\) -> \(76\)- our last 2 digits
B that is
