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S_A_V [24]
2 years ago
11

John took his pennies out of his piggy bank and discovered that he could put his pennies in equal piles of three, four, five, si

x, seven or eight. What is the least number of pennies that he could have had?
Mathematics
2 answers:
alexgriva [62]2 years ago
3 0

Answer:

Least number of pennies is 840


Step-by-step explanation:

To get the least number of pennies, we need to figure out the LCM of 3, 4, 5, 6, 7, & 8.


We do prime factorization of each number and take each unique digit the most number of times it occurs in any one of them.


3 is 3

4 is 2*2

5 is 5

6 is 3*2

7 is 7

8 is 2*2*2


There are 2s, 3s, 5s, and 7s. Two occurs three times (highest) in 8, so we take 2 three times, 3,4,5,&7 all occur highest 1 time. So we take them one.


So LCM = 2*2*2*3*5*7 = 840

Murrr4er [49]2 years ago
3 0
The answer is 840 I think
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Step-by-step explanation:

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Step-by-step explanation:

l = 4w

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2 years ago
2 and 2 over 2, divided by -1/4 = what
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3 years ago
What is 30% as a fraction
krok68 [10]
 = 30%

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3 years ago
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One of our brainliest, Konrad509, made this:
Reil [10]

\dfrac{B_x \sqrt{74_x}}{1D_x}+J_x51_x=4G3_x

A=10, B=11, C=12, etc.

\dfrac{11\cdot x^0\cdot \sqrt{7\cdot x^1+4\cdot x^0}}{1\cdot x^1+13\cdot x^0}+19\cdot x^0\cdot (5\cdot x^1+1\cdot x^0)=4\cdot x^2+16\cdot x^1+3\cdot x^0\\\\\dfrac{11\sqrt{7x+4}}{x+13}+19(5x+1)=4x^2+16x+3\\\\\dfrac{11\sqrt{7x+4}}{x+13}+95x+19=4x^2+16x+3\\\\11\sqrt{7x+4}+95x(x+13)+19(x+13)=(4x^2+16x+3)(x+13)\\\\11\sqrt{7x+4}+95x^2+1235x+19x+247=4x^3+52x^2+16x^2+208x+3x+39\\\\11\sqrt{7x+4}=4x^3-27x^2-1043x-208\\\\121(7x+4)=(4x^3-27x^2-1043x-208)^2

121(7x+4)=(4x^3-27x^2-1043x-208)^2\\\\847x+484=16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433888 x + 43264\\\\16 x^6 - 216 x^5 - 7615 x^4 + 54658 x^3 + 1099081 x^2 + 433041 x +42780=0

Now, the "only" thing that remains to do is solving the above equation.

While making this problem I only made sure it has a solution. I didn't try to solve it myself and I didn't know it will end up with such "convoluted" polynomial. Sorry to everyone who tried to solve it... m(_ _)m

I think the best way to approach it is using the rational root theorem since we know that x\in\mathbb{N}. Moreover we can deduce that x\geq19 since there is J and J=19.

After you succesfully solve it, you should get the answer x=20.

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