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22 Nov 2021, 09:35
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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. \(30\)
A. \(18\)
B. \(20\)
C. \(22\)
D. \(25\)
E. \(30\)
22 Nov 2021, 09:35
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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 \(n^2=xywz\) ).
Thus, \(n-m=11\) 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
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 \(n^2=xywz\) ).
Thus, \(n-m=11\) 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
