Mathematics, applications of mathematics to life in general, and my life as a mathematician.
Puzzle. Find the largest and the smallest 4-digit numbers n such that when you erase the first two digits of n, you get the sum of the digits of n.
9927 and 1818
Very clever! Those values look right, but can you prove that they are the largest and smallest?
a + b + c + d – 10c + d leads to the necessary condition that a + b = 9c
This fits your numbers but I cannot go any further to eliminate smaller and larger four-digit numbers.
9929 and 1810
Brute force is possible, using a small python program to browse through all numbers between 1000 and 9999.
I get the same answer as Leo.
We have n = 1000a + 100b + 10c + d, where 0 <= b, c, d <= 9 and 1 <= a <= 9 (because n is a four-digit number), and a+b+c+d = 10c+d (the sum of all four digits is equal to the result of erasing the first two digits). From a+b+c+d = 10c+d, subtracting c+d from both sides, we get a+b = 9c. Since 1<=a<=9 and 0<=b<=9 we have 1<=a+b<=18, so the only possibilities for c are 1 (giving a+b=9) and 2 (giving a+b=18). In case c=1, where we have b=9-a, the largest possible n comes from choosing the largest possible a and d, namely a=d=9, and then b=9-a=0, so we get 9019. The smallest possible n comes from choosing the smallest possible a and d, namely a=1 and d=0, and then b=9-a=8, giving 1810. In case c=2, where we have a+b=19, both a and b must be 9. We only get to choose d, so the largest possible n in this case is 9929 and the smallest is 9920.
So overall, as others have already said, the largest possible n is 9927 (with 9+9+2+7 = 27) and the smallest is 1810 (with 1+8+1+0 = 10).
Oops! *”where we have a+b=18″, not 19!
Oops again: *”the largest possible n is 9929 (with 9+9+2+9 = 29)”.
(Moderator — would you care to fix those two mistakes in my original reply, to tidy things up?)
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