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And Yet Another Question . . .

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The five dight number 9abc2 is divisible by 11, where a, b and c are diffrent one dight whole numbers. How many combinations of dights are possible values of a,b and c?

Jan 20, 2019
edited by Dominator416  Jan 20, 2019

#1
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this means that 9-a+b-c+2 is divisible by 11, so b-(a+c) is divisible by 11.

so this value is either -11 or 0.

if it is -11, we can use casework.

first take the case where b is 0.

the a can be 2, c can be 9, or a can be 3, and c can be 8, all the way up to the case where a is 9 and c is 2.

this has 8 possibilities.

now you take the case where b is 1. the sums are 12, and there are 7 posibilities using the same logic.

you keep doing this untill you get 1 possibility, so there are 8+7+6+5+...+1 possibilities or 28.

now if b-(a+c)=0, then it is the same.

if b is 0, there is one possibility.

if b is 1 there are 2 possibilities.

and you keep doing this until b is 9 and there are 10 possibilities.

this makes 1+2+3+...+10=55 possibilities for that case.

now you just add them up and get 55+28=83 possibilities.

HOPE THIS HELPED!!

Jan 20, 2019
#2
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$$\text{You start your answer with an assertion that needs proving}\\ 9abc2 \equiv 0 \pmod{11} \Rightarrow (9-a+b+c+2) \equiv 0 \pmod{11}$$

$$90000 + 1000a + 100b+10c+2 \equiv 0 \pmod{11}\\ 9 + 10a + b +10c+2 \pmod{11} = \\ 11-a+b-c \pmod{11} = \\ b - (a+c) \pmod{11}$$

Neat little trick.

Rom  Jan 20, 2019
#3
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asdf335:    I think you have made several mistakes in your reasoning.

1 - 8 + 7 + 6 + 5 +........+ 1 = 36 not 28

2- You should remove all numbers beginning with zeros to the right of 9, because they repeat: E.G. 9 011 2, 9 022 2, 9 033 2.......etc. There are 9 in total.

3- For each subsequent number beginning with 1, there are about 2 numbers repeated, which should also be removed: E.G. 9 112 2, 9 122, 2. There are 2 such numbers in each category beginning with 1 and ending in 9.  The only exceptions are the numbers beginning with 5: 95062, 95172, 95282, 95392, 95502, 95612, 95722, 95832, 95942. Out of these 9 numbers, there is one repeat, i.e,: 9 550 2, which should be removed.

4- There are a total of 9 x 10 =90 - 9 zeros - (2 x 9) + 1 for the other 9 numbers = 64 combinations of the 3 numbers a, b, c.

5- Here are All the numbers that meet the condition that they be "distinct", or different from each other:

91322, 91432, 91542, 91652, 91762, 91872, 91982, 92092, 92312, 92532, 92642, 92752, 92862, 92972, 93082, 93192, 93412, 93522, 93742, 93852, 93962, 94072, 94182, 94292, 94512, 94622, 94732, 94952, 95062, 95172, 95282, 95392, 95612, 95722, 95832, 95942, 96052, 96272, 96382, 96492, 96712, 96822, 96932, 97042, 97152, 97262, 97482, 97592, 97812, 97922, 98032, 98142, 98252, 98362, 98472, 98692, 98912, 99022, 99132, 99242, 99352, 99462, 99572,  99682=64

Jan 20, 2019
edited by Guest  Jan 20, 2019