For Bunuel/ every body ,Buzuets id

Question

I cant mail anyone so just to catch people's attention , i have a catchy subjectline
hey bunuel u have any idea about Buzuet's Identity .Please explain .All my mails are stuck 
I found the concept at  https://totalgadha.com/mod/forum/discuss.php?d=6959 
Bézout's  identity is a linear diophantine equation. It states that if a and b are nonzero integers with greatest common divisor d, then there exist integers x and y such that ax + by = d. Additionally, d is the smallest positive integer for which there are integer solutions x and y for the preceding equation.

Note that x and y can be negative also. If we take x and y to be positive numbers, then we can find a number such that all the numbers above that can be written as a linear combination of a and b.

On a dartboard, the only two points you can score are 9 and 16. What is the highest score you cannot obtain through this dartboard? 
A)121
B)123
C)117
D119
E)118
Answer: A number n will give remainders 0, 1, 2, 3, 4, 5, 6, 7, and 8 with 9. If it gives 0 (number is 9k) we only need some number of 9s to make up the number. If it gives remainder with 9 as

1 (9k + 1) we need 5 16s (9k + 1 + 80) to get remainder by 9 as zero.
2 (9k + 2) we need 1 16 (9k + 2 + 16) to get remainder by 9 as zero.
3 we need 6 16s to get remainder by 9 as zero.
4 we need 2 16s to get remainder by 9 as zero.
5 we need 7 16s to get remainder by 9 as zero.
6 we need 3 16s to get remainder by 9 as zero.
7 we need 8 16s to get remainder by 9 as zero.
8 we need 4 16s to get remainder by 9 as zero. _________________
For every kudos u give me , u get 1

Bunuel · Answer

First of all, there is nothing wrong with mailing service. Mail will remain in outbox till a recipient reads it. Next, you won't need Bézout's identity for GMAT. Diophantine equations (equations whose solutions must be integers only) you'll encounter on GMAT can and should be solved by trial and error. Check this for examples: gmat-prep2-92785.html?hilit=linear%20type Hope it helps.

Zatarra · Answer

Hi Bunuel i have 7 posts stuck in my mail box .so, i thiught of this way . Yea i know its not there but i just thought that since u r a math wixz u might be knowing this concept

Bunuel · Answer

I'm aware of it but I've never used it when solving GMAT problems, so wouldn't worry about it at all.

gmat1011 · Answer

ha ha - i looved this conversation

Munda Singh - i like your enthusiasm
Bunuel - i like how you keep it simple, no-nonsense and to the point! i have a sneaky feeling you'll ace the gmat if you were (is subjunctive mood correct here?) to ever try...

Zatarra · Answer

On a lighter note, My enthu is so high , had i a trampoline i would have broken it by jumping , but still i wudnt have come up to bunuels level.

144144 · Answer

Im not sure i got it.

Can someone plz give a full detailed explanation?

thanks.

shrouded1 · Answer

Consider all the possible numbers that we can form using 9 & 16. To make this easier split them into different sets, according to the remainder they leave with 9 :

1) Remainder 0 : These are divisible by 9. And all of these can be formed by using 9 alone
2) Remainder 1 : The number is of the form 9x+1. We need 9x+1=9a+16b where a & b are non-negative integers. In other words we need, 16b=9(x-a)+1. The smallest b for which this happens is 4. 16x4=64=9x7+1
Hence, the smallest such number possible is 64, then 64+9, 64+18, 64+27, ... and so on 
So all numbers of the form 9k+1, greater than or equal to 64 can be formed by 9s and 16s
3) Remainder 2 : Similar logic applies ... smallest number is 128
4) Remainder 3 : Smallest number is 48
5) Remainder 4 : Smallest number is 112
6) Remainder 5 : Smallest number is 32
7) Remainder 6 : Smallest number is 96
8) Remainder 7 : Smallest number is 16
9) Remainder 8 : Smallest number is 80

From all these series, the series that starts in the last is remainder 128 (9k+2)
So the previous number in this series must be the greatest number which we cannot form using 9 & 16 ... 128-9=119

For Bunuel/ every body ,Buzuets id

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