find the number of three digit numbers that leave remainder 9 when divided by 11

Question

shane_b · Answer

Basically, your first number will be 108...and then every number +11 after that up to 999:

108, 119, 130, 141, 152, 163, 174, 185, 196, 207, 218, 229, 240, 251, 262, 273, 284, 295, 306, 317, 328, 339, 350, 361, 372, 383, 394, 405, 416, 427, 438, 449, 460, 471, 482, 493, 504, 515, 526, 537, 548, 559, 570, 581, 592, 603, 614, 625, 636, 647, 658, 669, 680, 691, 702, 713, 724, 735, 746, 757, 768, 779, 790, 801, 812, 823, 834, 845, 856, 867, 878, 889, 900, 911, 922, 933, 944, 955, 966, 977, 988

shane_b · Answer

BTW: That's 81 values in all that meet that criteria.

anonymous · Answer

1st 3 digit num wch is divisible by 11 is 110 den 121 den 132 n so on..
for remindr 9 u need to add 9 in those numbers..
so 1st 3 digit num wch is divisible by 11 is 110 and last num is 990.
990=110+(n-1)11
ull get value of n=81
if u add 9 to those num,ull get 81...
108 is also a num wch is divisible by 11 nd reminder is 9...
so total num's will b 82.
and is 82

ganeshie8 · Answer

yh.. @arch1 is correct. 

numbers that leave 9 when divided by 11 are in from : 11x + 9
x = [9, 90] satisfies above expression for 3 digit numbers

total 82 numbers that match the constraint...

shane_b · Answer

Yes, I mistakenly left out 999...so it's 82.