Math  /  Numbers & Operations

Question8. Solve each of the puzzle-problems below: (e) Euler, 1770 . Divide 100 into two summands such that one is divisible by 7 and the other by 11 .

Studdy Solution
Test values of a a :
- For a=2 a = 2 :
1007(2)=86 100 - 7(2) = 86 b=8611=7.818 b = \frac{86}{11} = 7.818 (not an integer)
- For a=3 a = 3 :
1007(3)=79 100 - 7(3) = 79 b=7911=7.182 b = \frac{79}{11} = 7.182 (not an integer)
- For a=4 a = 4 :
1007(4)=72 100 - 7(4) = 72 b=7211=6.545 b = \frac{72}{11} = 6.545 (not an integer)
- For a=5 a = 5 :
1007(5)=65 100 - 7(5) = 65 b=6511=5.909 b = \frac{65}{11} = 5.909 (not an integer)
- For a=6 a = 6 :
1007(6)=58 100 - 7(6) = 58 b=5811=5.273 b = \frac{58}{11} = 5.273 (not an integer)
- For a=7 a = 7 :
1007(7)=51 100 - 7(7) = 51 b=5111=4.636 b = \frac{51}{11} = 4.636 (not an integer)
- For a=8 a = 8 :
1007(8)=44 100 - 7(8) = 44 b=4411=4 b = \frac{44}{11} = 4 (integer)
Thus, a=8 a = 8 and b=4 b = 4 is a solution.
The two summands are:
x=7×8=56 x = 7 \times 8 = 56 y=11×4=44 y = 11 \times 4 = 44
Therefore, the two numbers are 56 \boxed{56} and 44 \boxed{44} .

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