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The Question of the Day – 15 points
Stella sells 30 gismos a month at a price of 20 stickers per gismo. She notices that each time she raises the price of gismo by 4 stickers, she sells 2 few gismos per month. What is the maximum amount of revenue, in stickers, that Stella can make in a month of selling gismos?
The number of possible numbers was 6. The possible numbers were 5693, 5783, 5873, 5963, 7961 and 7691.
Working it out:
Bob's four-digit number can be written as 1000x+100y+10z+t. When he reverses the order of the digits, the sum of the two numbers can be written as 1001x+110y+110z+1001t. This can be factorised as 11(91x+10y+10z+91t). Because this number is divisible by 11, the perfect cube it is 990 less than must be divisible by 11 also, because 990 is divisible by 11. This narrows the possible perfect cubes down to 11, 22, 33, 44, etc. However, 11^3 would be too small and 33^3 and 44^3 are divisible by 3 and 16 respectively. Multiples of 11 higher than 44, when cubed, are far too large. Therefore, the perfect cube the number is 990 less than must be 22^3, or 10,648. Now we must find sets of 1-digit integers x, y, z and t that satisfy the equation:
Subtracting 990 from both sides, we get 11(91x+10y+10z+91t)=9,658.
Dividing both sides by 11, we get 91x+10y+10z+91t=878.
Because 10y+10z will not change the units digit, the units digit of 91t+91x must equal the units digit of 878, which is 8, or:
91(t+x)≡8 mod 10.
We can change this to t+x≡8 mod 10, because multiplying something by a number with a units digit of 1 will not change the units digit.
From this, we can conclude that t+x=8 or t+x=18, because no two single-digit numbers can add up to something more than 18.
However, because all the numbers in Bob's original number were different, and because the only way we can get 18 by adding two single digit numbers together is 9+9, t+x must equal 8. We will deal with the separate values of t and x later.
Because t+x=8, 91t+91x=728.
Substituting this value in the equation 91x+10y+10z+91t=878, we get:
728+10y+10z=878. Subtracting 728 from both sides, we get:
10y+10z=150, and dividing both sides by 10, we get:
We have concluded that t+x=8 and y+z=15.
There are 4 possible values of y. It can be 6,7,8 or 9.
The first digit of Bob's original number was prime and odd. let us try for the first odd prime- 3.
If x=3, then t=5 because x+t=5. However, Bob's original number was not divisible by 5, so that is not a solution.
If x=5, then t=3. This is a valid solution.
Let us try and find y and z now. Y can be 6, 7, 8 and 9 and z is 15-y. Therefore, the possible solutions with 5 as the first digit are as follows: 5,693, 5,783, 5,873 and 5,963.
If x=7, then t=1. This is also a valid solution.
Let us try and find y and x now. Y can be 6 and 9 because if y is 7 or 8, we will repeat digits. Therefore, the possible solutions with 7 as the first digit are 7,691 and 7,961.
There are no more solutions because x cannot be more than 8 because, if so, t would be a negative number. Counting the solutions, we find that we have found 6 solutions in total. ∎
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