Difference between revisions of "1987 AJHSME Problems/Problem 8"

Revision as of 18:49, 30 May 2017

If $\text{A}$ and $\text{B}$ are nonzero digits, then the number of digits (not necessarily different) in the sum of the three whole numbers is

$\begin{tabular}[t]{cccc} 9 & 8 & 7 & 6 \\ & A & 3 & 2 \\ & & B & 1 \\ \hline \end{tabular}$

$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ \text{depends on the values of A and B}$

The minimum possible value of this sum is when $A=B=1$ , which is $9876+132+11=10019$

The largest possible value of the sum is when $A=B=9$ , making the sum $9876+932+91=10899$

Since all the possible sums are between $10019$ and $10899$ , they must have $5$ digits.

$\boxed{\text{B}}$

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

@@ Line 16: / Line 16: @@
 The minimum possible value of this sum is when <math>A=B=1</math>, which is <cmath>9876+132+11=10019</cmath>
-The largest possible value of the sum is when <math>A=B=9</math>, making the sum <cmath>9876+999+91=10966</cmath>
+The largest possible value of the sum is when <math>A=B=9</math>, making the sum <cmath>9876+932+91=10899</cmath>
-Since all the possible sums are between <math>10019</math> and <math>10966</math>, they must have <math>5</math> digits.
+Since all the possible sums are between <math>10019</math> and <math>10899</math>, they must have <math>5</math> digits.
 <math>\boxed{\text{B}}</math>