1998 AHSME Problem 2

Number Theory Level 1

Letters A,B,C, and D represent four different digits selected from 0, 1, 2, ... , 9. If $\frac{ A+B}{C+D}$ is an integer that is as large as possible, what is the value of $A+B$ ?

15 16 14 17

3 solutions

Piolo Andrew Azuela
Mar 22, 2014

Simple logic...

A = 9

B = 8

A and B being the 2 largest numbers in the set

C = 0

D = 1

C and D being the 2 smallest numbers in the set

(C+D) = 1, so any number can be placed in replacement of A and B other than 0 and 1

Then, the highest possible is needed so the 2 largest numbers will replace A and B, numbers 8 and 9

So (A+B) = 17

(8+9) / (0+1) = 17

$\boxed{17}$ is the answer

The numerator is independent of the integer (practically), since the largest numbers will go in the numerator. You dont need to determine the denominator, or do any division.

Star Light - 7 years, 2 months ago

Ashish Menon
May 31, 2016

The denominator can have $0$ and $1$ as $C$ and $D$ . So, put $A + B = 8 + 9 = \color{#69047E}{\boxed{17}}$ to maximize the value.

Hmm super duper easy question

Rishabh Sood - 4 years, 11 months ago

Yep hehe :)

Ashish Menon - 4 years, 11 months ago

Mohd Naim Mohd Amin
Apr 7, 2014

Hey yo pals,

As the selected range of A,B,C & D are [0.9],

as (A+B) / (C+D) = maximum possible value,

C+D = the lowest value, since for the lowest value of C+D=1,(C=0, D=1 or D=0, C=1),

A+ B / 1 = maximum possible value,

A + B = 8 + 9 = 17 as A = 8, B=9 or A=9, B=8 ,

therefore, maximum possible value = 17,

thanks.....

1998 AHSME Problem 2

3 solutions

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