101 for 100
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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3 solutions
How did you get that Only whole numbers for x and y are 3 and 11?
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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 4 b ≡ 1 ( m o d 1 1 ) , which means b ≡ 4 1 ( m o d 1 1 ) . Since 4 1 ≡ 3 ( m o d 1 1 ) , 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. : )
By substituting all values from 1 to 8 for x in the last equation and finding whole number values for y.
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.
Why did you minus 1 to get rid of the z? and how did you get the 199?
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You have to eliminate one variable. I picked z.
I got 199 by subtracting 101 from 300.
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Oh I see. For some reason I didn't see the x + y + z over top of #2.
300 - 101 = 199
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
x + y + z = 1 0 1 . . . . ( 1 ) 9 x + 4 y + ( 3 1 ) z = 1 0 0 2 7 x + 1 2 y + z = 3 0 0 . . . . ( 2 )
( 2 ) − ( 1 ) = 2 6 x + 1 1 y = 1 9 9
2 6 x ≡ 1 9 9 ( m o d ( − 1 1 ) ) ⟺ 4 x ≡ 1 ( m o d ( − 1 1 ) ) ⟺ 4 x − 1 1 ≡ 1 ( m o d ( − 1 1 ) ) ⟺ 4 x ≡ 1 2 ( m o d ( − 1 1 ) ) ⟺ x ≡ 3 ( m o d ( − 1 1 ) ) x = 3 − 1 1 k → x is positive integer only if k=0 so, x = 3
2 6 x + 1 1 y = 1 9 9
2 6 ∗ 3 + 1 1 y = 1 9 9
x + y + z = 1 0 1 . . . . ( 1 ) 9 x + 4 y + ( 3 1 ) z = 1 0 0 2 7 x + 1 2 y + z = 3 0 0 . . . . ( 2 )
( 2 ) − ( 1 ) = 2 6 x + 1 1 y = 1 9 9
2 6 x ≡ 1 9 9 ( m o d ( − 1 1 ) ) ⟺ 4 x ≡ 1 ( m o d ( − 1 1 ) ) ⟺ 4 x − 1 1 ≡ 1 ( m o d ( − 1 1 ) ) ⟺ 4 x ≡ 1 2 ( m o d ( − 1 1 ) ) ⟺ x ≡ 3 ( m o d ( − 1 1 ) ) x = 3 − 1 1 k → x is positive integer only if k=0 so, x = 3
2 6 x + 1 1 y = 1 9 9
2 6 ∗ 3 + 1 1 y = 1 9 9
⟺ y = 1 1
so, z = 1 0 1 − ( 3 + 1 1 )
z = 8 7
( 1 ) : ( 2 ) : ( 2 ) × 3 = ( 3 ) : ( 3 ) − ( 1 ) : x + y + z = 1 0 1 9 x + 4 y + 3 1 z = 1 0 0 2 7 x + 1 2 y + z = 3 0 0 2 6 x + 1 1 y = 1 9 9
The only possible whole numbers for x and y are 3 and 1 1 .
Therefore the numbers of peanuts is 8 7 .