achloes wrote:
ScottTargetTestPrep BrushMyQuant Bunuel chetan2u yashikaaggarwalI was able to arrive at the answer without breaking apart the denominator 10 into 2 x 5.
However, when I do break it apart: (2^15 x 3^16 x 7^16) produces units digits of 6, and 1 and 1. But dividing the product by the denominator 5 produces a remainder of 1.
Could anyone help me understand why this method doesn't work?
Hey
achloesWhile other experts whom you've tagged to the post respond, sharing my two cents if that helps.
Before we figure out the reason why you're seeing a difference in answer let's take a general example to understand the concept -
What's the remainder when 10 is divided by 12 ?
The remainder (\(\frac{10}{12}\)) = 10, we can reason this well enough using logic right.
Now, let's say we divide the numerator and the denominator by 2
Remainder (\(\frac{10}{12}\)) = Remainder (\(\frac{5}{6}\)) = 5
Is the remainder when 10 divided by 12, 5? Of course, not!
What's the difference, the remainder got divided by 2 as well. In the first case, the remainder is 10, and in the second case, the remainder is 5.
The general rule is: Whenever you divide the numerator and denominator by a common factor, the common factor has to be multiplied (accounted for) to obtain the net remainder.
In our example, the common factor is 2.
Remainder (\(\frac{10}{12}\)) = Remainder (\(\frac{5}{6}\)) = 5
The final remainder = 5 * 2 = 10. This is what we're getting in the earlier example.
Now let's extend this concept to the question -
What is the remainder when \((2^{16})(3^{16})(7^{16})\) is divided by 10 ?
If we find the unit's digit and use the concept of cyclicity we will get the unit digit of the product \((2^{16})(3^{16})(7^{16})\) = 6 and that's our reminder. I guess you were able to get the answer without cancellation.
Now, let's see what happens when you cancel 2 from the numerator and denominator
Remainder (\(\frac{2^{16} * 3^{16}*7^{16}}{10}\))
= Remainder (\(\frac{2^{15}*3^{16}*7^{16}}{5}\))
Using the concept of cyclicity, we can find the units' digits
= Remainder (\(\frac{8* 1 * 1}{5}\)) = 3
Just as in the previous example, the remainders are no longer the same. In fact, the new remainder is half of the original remainder and this is expected
So to get the actual remainder, we need to multiply the obtained value by the factor used, i.e. multiply the new remainder by 2 to get the actual remainder.
3 * 2 = 6
Now by both methods, the remainders are the same.
Hope this help.
_________________
Want to discuss quant questions and strategies : Join the quant chat group todayPower of Tiny Gains1.01^(365) = 37.780.99^(365) = 0.03