Two families lived next door to each other —the Abramowitzs and the Bernhardts. The total ages of the four members of the Abramowitz family amounted to one hundred years, and the total ages of the four members of the Bernhardt family also amounted to the same.
It was found in the case of each family that the sum obtained by adding the squares of each of the children's ages to the square of the mother's age equaled the square of the father's age. In the case of the Abramowitz family, however, Emmylou was one year older than her brother Eric, whereas Francine Bernhardt was two years older than her brother Frank.
What was the age of each of the eight individuals?
(In reply to
re: Computer Assisted Solution (spoiler) by Dej Mar)
i wrote a few lines of code to brute force the answer out. here's the what i got:
dadA's age: 39
momA's age: 34
girlA's age: 14
boysA's age: 13
dadB's age: 42
momB's age: 40
girlB's age: 10
boysB's age: 8
and here's the C source code :
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void main() {
for (int dA = 1; dA < 100; dA++) {
for(int mA = 1; mA < 100; mA++) {
for (int bA = 1; bA < 100; bA++) {
if (dA + mA + 2*bA == 99) {
if (pow(dA, 2) == pow(mA, 2) + 2*pow(bA, 2) + 2*bA + 1) {
printf("dadA's age: %d\n", dA);
printf("momA's age: %d\n", mA);
printf("girlA's age: %d\n", bA + 1);
printf("boysA's age: %d\n", bA);
}
}
}
}
}
printf("\n");
for (int dB = 1; dB < 100; dB++) {
for(int mB = 1; mB < 100; mB++) {
for (int bB = 1; bB < 100; bB++) {
if (dB + mB + 2*bB == 98) {
if (pow(dB, 2) == pow(mB, 2) + 2*pow(bB, 2) + 4*bB + 4) {
printf("dadB's age: %d\n", dB);
printf("momB's age: %d\n", mB);
printf("girlB's age: %d\n", bB + 2);
printf("boysB's age: %d\n", bB);
}
}
}
}
}
}
computer solved it for me
