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Eleanor is making chocolate chip cookies for her friends. If she divides the cookies equally among \(11\) of her friends, she'll have \(4\) cookies left over. If she divides the cookies equally among \(7\) of her friends, she'll have \(1\) cookie left over. Assuming that Eleanor made fewer than \(100\) cookies, what is the sum of the possible numbers of cookies that she could have made?

Eleanor is making chocolate chip cookies for her friends. If she divides the cookies equally among \$11\$ of her friends, she'll have \$4\$ cookies left over. If she divides the cookies equally among \$7\$ of her friends, she'll have \$1\$ cookie left over. Assuming that Eleanor made fewer than \$100\$ cookies, what is the sum of the possible numbers of cookies that she could have made?

So I got 15 as one solution but that was by process of elimination.  Is there a faster way?

Dec 22, 2018

#1
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I would take into consideration the LCM of [7, 11] =77

So, 77n + 15, where n =0, 1, 2, 3.......etc.

Dec 22, 2018

#1
+2

I would take into consideration the LCM of [7, 11] =77

So, 77n + 15, where n =0, 1, 2, 3.......etc.

Guest Dec 22, 2018
#2
+812
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Let's name X as the number of cookies she has.

Y is the number of cookies she gives to each of her 11 friends, and Z us the number of cookies she gives to each of her 7 friends.

We have X = 11Y + 4 and X = 7Z + 1, and by combining the equations, we have 11Y + 4 = 7Z + 1. Subtracting 4 on both sides, we have 11Y = 7Z - 3, and by dividing both sides by 11, we have \(Y = \dfrac{7Z-3}{11}\). Follow through with the equation.

Dec 22, 2018
edited by PartialMathematician  Dec 22, 2018