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Difference between revisions of "1953 AHSME Problems/Problem 24"

(Created page with "If <math>a,b,c</math> are positive integers less than <math>10</math>, then <math>(10a + b)(10a + c) = 100a(a + 1) + bc</math> if: <math>\textbf{(A)}</math> <math>b+c=10</mat...")
 
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==Problem==
 
If <math>a,b,c</math> are positive integers less than <math>10</math>, then <math>(10a + b)(10a + c) = 100a(a + 1) + bc</math> if:
 
If <math>a,b,c</math> are positive integers less than <math>10</math>, then <math>(10a + b)(10a + c) = 100a(a + 1) + bc</math> if:
  
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<math>\textbf{(A)}</math> <math>b+c=10</math>
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<math>\textbf{(A) }b+c=10</math>
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<math>\textbf{(B)}</math> <math>b=c</math>
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<math>\qquad\textbf{(B) }b=c</math>
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<math>\textbf{(C)}</math> <math>a+b=10</math>
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<math>\qquad\textbf{(C) }a+b=10</math>
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<math>\textbf {(D)}</math> <math>a=b</math>
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<math>\qquad\textbf {(D) }a=b</math>
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<math>\textbf{(E)}</math> <math>a+b+c=10</math>
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<math>\qquad\textbf{(E) }a+b+c=10</math>
  
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==Solution==
 +
Multiply out the LHS to get <math>100a^2+10ac+10ab+bc=100a(a+1)+bc</math>. Subtract <math>bc</math> and factor to get <math>10a(10a+b+c)=10a(10a+10)</math>. Divide both sides by <math>10a</math> and then subtract <math>10a</math> to get <math>b+c=10</math>, giving an answer of <math>\boxed{A}</math>.
  
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Multiply out the LHS to get <math>100a^2+10ac+10ab+bc=100a(a+1)+bc</math>. Subtract <math>bc</math> and factor to get <math>10a(10a+b+c)=10a(10a+10)</math>. Divide both sides by <math>10a</math> and then subtract <math>10a</math> to get <math>b+c=10</math>, giving an answer of <math>\boxed{A}</math>.
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==See Also==
 +
{{AHSME 50p box|year=1953|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 00:53, 26 January 2020

Problem

If $a,b,c$ are positive integers less than $10$, then $(10a + b)(10a + c) = 100a(a + 1) + bc$ if:

$\textbf{(A) }b+c=10$ $\qquad\textbf{(B) }b=c$ $\qquad\textbf{(C) }a+b=10$ $\qquad\textbf {(D) }a=b$ $\qquad\textbf{(E) }a+b+c=10$

Solution

Multiply out the LHS to get $100a^2+10ac+10ab+bc=100a(a+1)+bc$. Subtract $bc$ and factor to get $10a(10a+b+c)=10a(10a+10)$. Divide both sides by $10a$ and then subtract $10a$ to get $b+c=10$, giving an answer of $\boxed{A}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions


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