## Problem Of The Week 35

**How many 3-d igit numbers are such that one of the digits is the average of the other two?**

**(A) 96 (B) 112 (C) 120 (D) 104 (E) 256**

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**How many 3-d igit numbers are such that one of the digits is the average of the other two?**

**(A) 96 (B) 112 (C) 120 (D) 104 (E) 256**

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Written by Implex

September 28, 2008 at 4:44 pm

Posted in Number Thoery, Problem of the week

Tagged with Combinatorics, Number Theory, Problem of the week

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96

(1,2,3),(2,3,4)….(7,8,9) 7 triplets

(1,3,5),(2,4,6)…(5,7,9) 5 triplets

(1,4,7),(2,5,8),(3,6,9) 3 triplets

(1,5,8) 1 triplet

Total possible triplets =16

Total possible no.s =16* 3! =96

riyaziqSeptember 29, 2008 at 8:01 am

Zero has not been considered above. So when you consider zero, then one of the other number should be twice of another. there are 4 possibilities and each has 4 ways..so adding 16 the answer is 96+16=112. Answer choice B)

Celebrating LifeSeptember 29, 2008 at 10:16 am

nice way…

(a+b)/2=c

consider the terms in ap and arrange 🙂

outtimedSeptember 29, 2008 at 11:09 am

@Celebrating Life

the case for numbers like 111,222,333………999

has not been considered…

9 cases more are added thus my answer appears to be(96+16+9)=121

would anyone like to correct me?

wolverineOctober 1, 2008 at 3:44 am

Yeah @wolverine

your solution counts for those aps when terms are equal

but this is an old amc problem

and they did not give option for counting with that approach

so the option 121 is not there

answer is right

so as the term distinct is not mentioned

we should also count for there 9!

Good Job

outtimedOctober 1, 2008 at 11:17 am

then i must say it was a flawed question…!!!

wolverineOctober 1, 2008 at 5:19 pm

yupp, it was ambiguity is generally avoided by amc

outtimedOctober 1, 2008 at 6:22 pm

16*3!+9+4*(3!-2!)=121

SpidermanOctober 2, 2008 at 6:28 pm