limestone wrote:Such a hard work for me to solve this problem. There're some techniques that I will use to solve this:
xxxab * 11 = xxxxcb ( where c is the unit digit of a+b)
For example 23*11 = 253; note that 5 = 2+3
or 478* 11 = xx58 ( where 5 is the unit digit of 7 + 8)
(a+b)(a-b) = a^2 - b^2
the product of xxxab * xxxcd wil have the same unit and tenth digit with that of ab*cd
For example : 1114 * 5678 = xxxx92 and 14*78 = xx92
xxx5 * xxx5 = xxxxx25
First, I devide them into 3 pairs:
22x31
44x27
37x43
22 x 31 = 2*11* 31 = 2* xx41 ( use rule 1 when multiply with 11)
= xx82 = 82 (rule 3)
44x27 = 4*11*27 = 4*xx97 = 4*97 = 4*(100-3) = 400 - 12 = 388 = 88
37x43 = (40-3)*(40+3) = 40^2 - 3^2 = 1600 - 9 = 1591 = 91
Now we have 82*88*91
82*88 = (85-3)(85+3) = 85^2 - 3^2 = xx25 - 9 = xx16 = 16
Now we have only 16*91 left
16*91 = 16*(100-10 +1) = 1600 - 160 +16 = xx40 + 16 = xx56 = 56
Then Pick D.
I want to know if anyone has a shorter approach. It took me more than 5 minutes to solve this.
Another solution I have thought of is :
xxxab* xxxcd = xxxxxxxef
Where ab*d + b*c*10 = xef
Hats off to such a hard work, limestone. Actually, the number formed by the last two digits to the right is same as the remainder when the product is divided by 100.
The remainder when 22 X 31 X 44 X 27 X 37 X 43 is divided by 100
= The remainder when 22 X 31 X 11 X 27 X 37 X 43 is divided by 25 {this division by 4 should be compensated in the end}
= The remainder when 22 X 6 X 11 X 2 X 12 X 18 is divided by 25
= The remainder when 132 X 22 X 216 is divided by 25
= The remainder when 7 X 22 X 16 is divided by 25
= The remainder when 154 X 16 is divided by 25
= The remainder when 4 X 16 or 64 is divided by 25
= 14 {don't forget to multiply this by 4}
Hence, the required two-digit number is [spoiler]
56.
D[/spoiler]
