reto wrote:
Harley1980 wrote:
Bunuel wrote:
X is a three-digit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?
(1) The hundreds digit of XY is 6.
(2) The tens digit of XY is 4.
Kudos for a correct solution.
We know that we have two numbers \(abc\) and \(cba\) and from statements we have information about tens and hundreds.
In this case (when only \(1\) and \(2\) can be numbers) we know that tens will be equal to \(ab + bc\) and hundreds will be equal to \(a^2+b^2+c^2\)
1)\(a^2+b^2+c^2 = 6\)
This possible only in variants when two of numbers are equal to \(1\) and other number is equal to \(2\)
\(1^2+1^2+2^2 = 6\) etc.
So \(abc\) can be \(112\), \(211\) or \(121\) All this numbers when divided by \(3\) give us remainder \(1\)
Sufficient
2) \(ab + bc = 4\)
This possible when \(ab\) and \(bc\) equal to \(2\). So we have two variants:
\(a\) and \(c\) equal to \(2\) and \(b\) equal to \(1\)
or \(a\) and \(c\) equal to \(1\) and \(b\) equal to \(2\)
\(212\) and \(121\). When divided by \(3\) first number gives remainder \(2\) and second number gives remainder \(1\)
Insufficient.
Answer is A
Dear Harley
I have never seen this rule before even i went through the
Manhattan Books - maybe I have overlooked it. However, is this rule with the tens and hundreds a general rule for that type of question?:
"We know that we have two numbers abc and cba and from statements we have information about tens and hundreds.
In this case (when only 1 and 2 can be numbers) we know that tens will be equal to ab+bc and hundreds will be equal to a2+b2+c2"
How does it behave if the question tells you that each digit can be 1, 2 or 3?
Thank you
Hi reto,
This rule is just based on how we multiply, there is nothing more to it.
X = abc = a*100 + b*10 + c
Y = cba = c*100 + b*10 + a
XY = (a*100 + b*10 + c)*(c*100 + b*10 + a)
=>XY = 10000*ac + 1000*ab + 100a^2 + 1000bc + 100b^2 + 10ab + 100c^2 + 10*bc + ac
=> XY = 10000*ac + 1000*(ab + bc) + 100*(a^2 + b^2 + c^2) + 10*(ab + bc) + ac
If it can be proved that ac, (ab+bc) and (a^2 + b^2 + c^2) are single digits then the unit digit will be ac, 10th place will be (ab+bc) and 100th place will be (a^2 + b^2 + c^2)
let's take the example of 112 and 211
112 * 211 = 23632
unit digit of 112 * 211 is ac = 1*2 = 2 < 10
10th digit of 112 * 211 is ab + bc = 1+2 = 3 < 10
100th digit 0f 112 * 211 is a^2 + b^2 + c^2 = 1 + 1 + 4 = 6 < 10
Now let's take the example 223 and 322
223*322 = 71806
ac = 2*3 = 6 < 10, so unit digit of 223*322 is 6
ab + bc = 4+6 = 10, so 10th digit of 223*322 is 0 not 10
a^2 + b^2 + c^2 = 4 + 4 + 9 = 17 > 10, so 100th digit of 223*322 is 7 + 1(carry over from the 10th digit) = 8 and not 17
Hope this helps. Do let me know if you have more questions.