GMAT CLUB OLYMPICS: Achilles's age is a 2-digit number ab and Hector's

Official Solution:

Achilles's age is a 2-digit number \(ab\) and Hector's age is a 2-digit number \(cd\). A 4-digit number \(abcd\) is the square of an integer. In 11 years, Achilles's age will be a 2-digit number \(xy\) and Hector's age will be a 2-digit number \(wz\). A 4-digit number \(xywz\) will also be the square of an integer. What is Achilles's age now?

A. \(18\)
B. \(20\)
C. \(22\)
D. \(25\)
E. \(30\)


\(xy=ab+11=10a+b+11\);

\(wz=cd+11=10c+d+11\);

\(abcd=1000a + 100b + 10c + d\);

\(xywz=100(10a + b + 11) + (10c+d+11)=(1000a + 100b + 10c + d) + 1111=abcd+1111\);

\(xywz - abcd=(abcd+1111)-abcd=1111\).

We are told that both \(abcd\) and \(xywz\) are perfect squares, say \(m^2\) and \(n^2\), respectively.

\(n^2-m^2=1111\);

\((n-m)(n+m)=1*1111=11*101\);

\(n-m=1\) and \(n+m=1111\) is not possible because in this case \(m = 555\) and \(n = 556\) but this numbers squared does not give a four-digit numbers (\(m^2=abcd\) and \(n^2=xywz\) );

\(n-m=11\) and \(n+m=101\). Solving gives \(m=45\) and \(n=56\);

\(abcd=m^2=45^2=2025\);

Therefore, Achilles's age now is \(ab=20\) years.


Answer: B
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IMO B.

I got it by number plugging.

I start by plugging A 18. Now xy = 29. Perfect square in 2900s is 2916. But notice then that wz = 16. Which means cd will NOT eb a two digit number. Eliminate.

Then I proceed to B.
Let ab = 20 (as in option B). xy will be 20+11 = 31.

The only perfect square I can think of in 3100's is 3136 which is square of 56.

So, wz = 36. Now, 36 - 11 = 25 = cd.

so abcd = 2025 which is again a perfect square of 9*5 = 45. (so option B satisfies all requirements).

So no need to go further than B

xy = ab + 11
wz = cd + 11

assume abcd = m^2, and xywz = n^2

xywz = (10a + b + 11) * 100 + 10c + d + 11 = 1000a + 100b + 10c + d + 1111

Notice that 1000a + 100b + 10c + d = abcd = m^2, so m^2 + 1111 = n^2 -> (n-m)(n+m) = 1111

Since m and n are both integers (as stated in the question), and 1111 = 11 * 101, n-m=11 and n+m=101, so n=56 and m=45. m^2 = 2025

So ab = 20, which is Achilles' age now. Notice that n-m=1 and n+m=1111 does not work because n=556 and m=555, the squares of which are beyond 4 digits.

So B).

abcd is a square (x^2) and

abcd + 1111 is a square (y^2)


y^2 - x^2 = 1111 = 11*101

y+x = 101
y-x =11

y=56 and x = 45

x^2 = 2025 and y^2 = 3136

B

Hello everyone

My answer is B

Here is my solution!

Say the number abcd is square of m and the other xywz is square of n. Also square of n is 1111 more than the square of m as we have increased each digit by 1
Let us see how the difference is 1111.
Given: After 11 years - ab becomes xy and cd becomes wz => xy=ab+11 and wz=cd+11
\(m^2=abcd = 100ab +cd\)
\(n^2=xywz = 100xy +wz = 100(ab+11) + cd+11 = 100ab+1100+cd+11 = 100ab+cd+1111\)
Thus \(n^2-m^2=(n-m)(n+m)=1111=11*101=1*1111\)

Only two cases
I. n-m=11 and n+m=101………
2n=112…..n=56 and m =45
Thus, abcd=\(45^2=2025\). ab=age of A now =20.

II. n-m=1 and n+m=1111………
2n=1112…..n=556 and m =555
NOT possible as squares of m and n will not be 4-digit numbers.


B
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Working with options works best for me in these type of questions.

Since it is guaranteed that the 4 digits with Achilles' current age is a perfect square, try to find the squares with the initial two digits mentioned in the options. For instance, the only 4-digit square beginning with 18 is 1849. Now we work with the next condition. Adding 11 to both numbers, the new 4-digit number will be 2960 which is not a perfect square. Hence, it is not the correct option.

Moving to next option we have 2025 as the square. Adding 11 we get, 3136 which is also a perfect square. As this satisfies both conditions this is our answer. Hence, current age of Achilles should be 20.

Range of both Achilles & Hector age has to be 10 to 88. Square root range from 32 to 99 to be less than 10,000.
Square of a number in this range when added with 1111 is also a square.
m is abcd and n is xywz
m^2+1111 = n^2 and still maintain n^2 less than 10,000.
The maximum value of n can be 56 as square root of 1111 is 33.333 approximately (obtained by substracting 89-33.333).

1000a+100b + 10c+d = m^2
1000(a+1) + 100(b+1)+10(c+1)+(d+1) = n^2

(d+1) cannot be greater than 6 as square of the double digit (say that it was 16) will be greater than 100.
(a+1) cannot be greater than 3 as square of the double digit (say that xy was 40) will be greater than 1000.(since we expect 4 digit number)
xy can be maximum 33
wz and cd should be a square of the number by itself.
It cannot be 2^2 as it is 4 which is less than double digit.
3^2 is again 9 which is less than double digit number.
So after incrementing the age's unit's and ten's digit (which is 11) , cd+11 = wz both should be square.
4^2, will have 16 as square, and if we add 11, it will be 27 which is not a square.
5^2, will have 25 as square, and if we add 11, it will be 36 which is a square as well, and matches all the condition.

=> cd = 25 and wz = 36.
Similarly we can prove that xy = 31 and hence ab = 20 ( as x cannot be greater than 3 and as the new age after adding 11 years to Achilles & Hector as xywz should be (56)^2 = 3136. This is of the form n^2 where n= 56.)
So when substracting 11 from the xy and wz, respectively we get abcd as 2025. This is of the form m^2 where m= 45.

So the age of Achilles currently is 20.

Therefore, B is the right choice.

Some perfect squares we know off the top of our head:

\(40^2 = 1600\)

\(45^2 = 2025\) (To get the square of a number ending in 5, we write 25 at the end and multiply the rest of the number by the consecutive integer. So 45^2 = (4*5) 25 = 2025

\(50^2 = 2500\)

\(55^2 = 3025\) (Write 25 first, then multiply 5 by 6 to get 30)

\(60^2 = 3600\)

\(65^5 = 4225\)

(A) Check 43^2 = 1849
But 2960 cannot be a perfect square because it ends with 1 zero.

(B) 2025 is a perfect square. Is 3136 a perfect square? It is if it is the square of 56. We see that it is.

(C) 47^2 = 2209 but 09 is not a two digit number. Ignore

(D) 2500 is a perfect square that doesn't work and next square will be more than 100 away i.e. 2601 so ignore.

(E) 55^2 = 3025. Is 4136 a perfect square? No because as we go forward, difference between adjacent perfect squares keeps increasing so the next perfect square will be more than 100 away.
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