Tom bought 101 nuts for $1.00. Brazil nuts were 9 cents each, walnuts were 4 cents each, and peanuts were 3 for 1 cent.

If he bought at least 1 of each type of nut, how many peanuts did he buy?

The answer is 87.

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How did you get that Only whole numbers for x and y are 3 and 11?

Ritu Roy
- 6 years, 8 months ago

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I found that 3 is the value for Brazil nuts by using moduli.

We know that 26b+11w=199. I wanted to find what value of b, modulus 11, would make it congruent to 199. 26b is congruent to 4b, and 199 is congruent to 1. So now, we have $4b \equiv 1 \pmod{11}$ , which means $b \equiv \frac{1}{4} \pmod{11}$ . Since $\frac{1}{4} \equiv 3 \pmod{11}$ , b=3, 14, 25, etc. b can only be 3 if w is positive, so there are indeed 3 Brazil nuts. From here, we can find that there are 11 walnuts and 87 peanuts.

Hope that was helpful if you wanted to know how to get the value for b without guessing and checking. : )

Rajiv Nelakanti
- 6 years, 8 months ago

By substituting all values from 1 to 8 for x in the last equation and finding whole number values for y.

Guiseppi Butel
- 6 years, 8 months ago

I used a system of equations: b+w+p=101 and 9b+4w+p/3=100. I used elimination method to eliminate p from the second equation and got 26b+11w=199. I repeatedly subtracted 26 from 199 until I was left with a multiple of 11. 199 - 3(26) = 121. Then w had to be 11, and substituting back into the original equation leaves 87 for p.

David Srebnick
- 6 years, 8 months ago

Why did you minus 1 to get rid of the z? and how did you get the 199?

A Former Brilliant Member
- 6 years, 8 months ago

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You have to eliminate one variable. I picked z.

I got 199 by subtracting 101 from 300.

Guiseppi Butel
- 6 years, 8 months ago

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Oh I see. For some reason I didn't see the x + y + z over top of #2.

A Former Brilliant Member
- 6 years, 8 months ago

300 - 101 = 199

Guiseppi Butel
- 2 years, 2 months ago

9b+4w+(p/3)=100 or 27b+12w+p=300 ---(i) & b+w+p= 101 ---(ii)

From (i) & (ii), 11p=15b+912 or p=(15b+912)/11 ---(iii)

To get whole number 'p' , b= 3, & p=87 ,so w=11

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$x+y+z = 101 .... (1)$ $9x+4y+ (\frac {1}3)z = 100$ $27x+12y+z = 300 ....(2)$

$(2) - (1)= 26x+11y = 199$

$26x \equiv 199 \pmod{(-11)}$ $\iff 4x \equiv 1 \pmod{(-11)}$ $\iff 4x-11 \equiv 1 \pmod{(-11)}$ $\iff 4x \equiv 12 \pmod{(-11)}$ $\iff x \equiv 3 \pmod{(-11)}$ $x = 3-11k \rightarrow$ x is positive integer only if k=0 so, $x=3$

$26x + 11y = 199$

$26*3 + 11y = 199$

$x+y+z = 101 .... (1)$ $9x+4y+ (\frac {1}3)z = 100$ $27x+12y+z = 300 ....(2)$

$(2) - (1)= 26x+11y = 199$

$26x \equiv 199 \pmod{(-11)}$ $\iff 4x \equiv 1 \pmod{(-11)}$ $\iff 4x-11 \equiv 1 \pmod{(-11)}$ $\iff 4x \equiv 12 \pmod{(-11)}$ $\iff x \equiv 3 \pmod{(-11)}$ $x = 3-11k \rightarrow$ x is positive integer only if k=0 so, $x=3$

$26x + 11y = 199$

$26*3 + 11y = 199$

$\iff y = 11$

so, $z = 101-(3+11)$

$\ \boxed{z=87}$

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$\begin{array}{rc} (1): & x + y + z = 101 \\ (2): & 9x + 4y + \frac{1}{3}z = 100 \\ (2) \times 3 = (3): & 27x + 12y + z = 300 \\ (3) - (1): & 26x + 11y = 199 \end{array}$

The only possible whole numbers for $x$ and $y$ are $3$ and $11$ .

Therefore the numbers of peanuts is $\boxed{87}$ .