Let 11 divide!

Number Theory Level 3

$\begin{aligned} \text{N}= {\color{#3D99F6}100a + 10b + c }\end{aligned}$ I have a three digits number as shown above which is always divisible by 11 where $a, b$ and $c$ are it's digits.

$\begin{aligned} & (1) \phantom{c}a \leq b , \phantom{c}a-b +c =0 \\& (2) \phantom{c} a >b, \phantom{c} a+b +c =11 \\& (3) \phantom{c} a > c ,\phantom{c} a-c -b =0 \\& (4) \phantom{c} c> a, \phantom{c} c-a-b = 0\\&\end{aligned}$

How many of the above statements are/is always true ?

4 1 All are false 3 All are true 2

1 solution

Giorgos K.
Feb 22, 2018

using Mathematica

case 1
#[[1]] - #[[2]] + #[[3]] & /@ Select[IntegerDigits /@ Table[11 i, {i, 10, 90}], #[[1]] <= #[[2]] &]

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} True

case 2
#[[1]] + #[[2]] + #[[3]] & /@ Select[IntegerDigits /@ Table[11 i, {i, 10, 90}], #[[1]] > #[[2]] &]

{11, 11, 13, 11, 13, 15, 11, 13, 15, 17, 11, 13, 15, 17, 19, 11, 13, 15, 17, 19, 21, 11, 13, 15, 17, 19, 21, 23, 11, 13, 15, 17, 19, 21, 23, 25} False

case 3
#[[1]] - #[[2]] - #[[3]] & /@ Select[IntegerDigits /@ Table[11 i, {i, 10, 90}], #[[1]] > #[[3]] &]

{0, 0, -2, 0, -2, -4, 0, -2, -4, -6, 0, -2, -4, -6, -8, 1, 0, -2, -4, -6, 3, 1, -1, 0, -2, -4, 5, 3, 1, -1, -3, 0, -2, 7, 5, 3, 1, -1, -3, -5, 0} False

case 4
#[[3]] - #[[1]] + #[[2]] & /@ Select[IntegerDigits /@ Table[11 i, {i, 10, 90}], #[[1]] < #[[3]] &]

{4, 6, 8, 10, 12, 14, 16, 7, 6, 8, 10, 12, 14, 5, 7, 8, 10, 12, 3, 5, 7, 10, 1, 3, 5, 7, 3, 5, 7, 5, 7, 7} False

Let 11 divide!

1 solution

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