A seven-digit positive integer is chosen at random. What is the probability that the number is divisible by 11, given that the sum of its digits is 61?
Answer Choices:
A. 143β
B. 113β
C. 72β
D. 114β
E. 73β
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Note that the maximum digit sum of a 7-digit number is 63. To compute how many 7-digit numbers have a digit sum of 63, we may think about it as distributing two β1's to the digits of 9999999. If they are distributed to the same digit, there are 7 ways to do so and if they are distributed to different digits there are (27β)=21 ways for a total of 28 numbers with digit sum 63.
A multiple of 11 must have an alternating digit sum of 11 as well, and note that 9999999 has an alternating digit sum of 9. Therefore, both β1's must be distributed to digit places which are negative - the 2nd, 4th, and 6th slots. Similarly to above there are 3+(23β)=6 ways to do this for a final answer of
286β=(A) 143ββ.
Suppose the 7-digit number was a1βa2βa3βa4βa5βa6βa7ββ. Then, let A=a1β+a3β+a5β+a7β and B=a2β+a4β+a6β. Note that Bβ€9β
3=27, and Aβ€9β
4=36. Thus, 34=61β27β€61βB=Aβ€36, and 25=61β36β€61βA=Bβ€27. Thus, 7β€AβBβ€11.
Note that if a1βa2βa3βa4βa5βa6βa7ββ is divisible by 11, AβB must also be divisible by 11, so AβB=11, implying A=36 and B=25. Note that (9βa2β)+(9βa4β)+(9βa6β)=27βB=2, so we can choose them in
(3β12+3β1β)=(24β)=6
ways for a2β,a4β,a6β. As A=36, a1β=a3β=a5β=a7β=9, so thus there are a total of 6 such 7-digit numbers divisible by 11 and sum 61.
To count the total number of 7-digit numbers with digit sum 61, we note that
k=1β7β(9βakβ)=63βk=1β7βakβ=2
so we can distribute them in
(7β12+7β1β)=(68β)=28
ways, for a probability of
286β=(A) 143ββ.
The problems on this page are the property of the MAA's American Mathematics Competitions