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Let Z be the largest number that is made up of each of the digits 1 through 9 exactly once and is
divisible by 99. What is the digit in the hundredth place, in such a number? Please note – A number
passes the test for 11 if the difference of the sums of alternating digits is divisible by 11.
Given : Z be the largest number that is made up of each of the digits 1 through 9 exactly once and is
divisible by 99.
To Find : digit in the hundredth place, in such a number
Solution:
Largest number possible = 987654321
Number should be divisible by 99 means 9 * 11
sum of 9 digits = 45 hence Divisible by 9
Divisible by 11
9 – 8 + 7 – 6 + 5 – 4 + 3 – 2 + 1
= ( 25) – (20)
= 5
we need to make this 5 as 0 or 11
it can not be made 0 as sum of all numbers = 45 Hence one sum will be odd and one will be even
so we nee to get 11
for that we need to get 28 – 17 = 11
Swap 1 with 4 or 3 with 6 or 5 with 8
but Swap 1 with 4 will give the largest number
Hence
987651324 is the largest is divisible by 99
9 – 8 + 7 – 6 + 5 – 1 + 3 – 2 + 4
= 28 – 17 = 11
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