The function $g$ is defined for each positive thre... | Free GMAT Practice Question | GMAT Panda

👋 Hi! I'm GMAT Panda. Ask me anything about this question, or click a button below to get started!

The function g is defined for each positive three-digit integer n by g (n) = 3^{a} \times 7^{b} \times 1 1^{c}, where a, b, and c are the hundreds, tens, and units digits of n, respectively. If s and t are three-digit positive integers such that g (s) = 21 g (t), then s - t =

Explanation	Calculations	Help
$We name the hundreds, tens, and units digits of s as a_{s}, b_{s}, and c_{s}, and those of t as a_{t}, b_{t}, and c_{t} .$	$g (s) = 3^{a_{s}} \times 7^{b_{s}} \times 1 1^{c_{s}}$
$We write the corresponding expression for g (t) .$	$g (t) = 3^{a_{t}} \times 7^{b_{t}} \times 1 1^{c_{t}}$
$We form the ratio of g (s) to g (t) .$	$\frac{g ( s )}{g ( t )} = 3^{a_{s} - a_{t}} \times 7^{b_{s} - b_{t}} \times 1 1^{c_{s} - c_{t}}$
$We set this ratio equal to 21 and express 21 as prime powers.$	$\frac{g ( s )}{g ( t )} = 21 = 3^{1} \times 7^{1}$
$We recognise that we can apply Prime Matching to match the exponents of corresponding primes.$	$a_{s} - a_{t} = 1$	Theory & Tactics Method Card ALG7 Click to view full details
$We match the exponent for the base 7.$	$b_{s} - b_{t} = 1$
$We match the exponent for the base 11.$	$c_{s} - c_{t} = 0$
$We express s - t in terms of the digit differences.$	$s - t = 100 \times (a_{s} - a_{t}) + 10 \times (b_{s} - b_{t}) + (c_{s} - c_{t})$
$We substitute the differences and calculate the result.$	$s - t = 100 \times 1 + 10 \times 1 + 0 = 110$

Scroll horizontally to view all columns

C

The function g is defined for each positive three-digit integer n by g (n) = 3^{a} \times 7^{b} \times 1 1^{c}, where a, b, and c are the hundreds, tens, and units digits of n, respectively. If s and t are three-digit positive integers such that g (s) = 21 g (t), then s - t =

Explanation	Calculations	Help
$We name the hundreds, tens, and units digits of s as a_{s}, b_{s}, and c_{s}, and those of t as a_{t}, b_{t}, and c_{t} .$	$g (s) = 3^{a_{s}} \times 7^{b_{s}} \times 1 1^{c_{s}}$
$We write the corresponding expression for g (t) .$	$g (t) = 3^{a_{t}} \times 7^{b_{t}} \times 1 1^{c_{t}}$
$We form the ratio of g (s) to g (t) .$	$\frac{g ( s )}{g ( t )} = 3^{a_{s} - a_{t}} \times 7^{b_{s} - b_{t}} \times 1 1^{c_{s} - c_{t}}$
$We set this ratio equal to 21 and express 21 as prime powers.$	$\frac{g ( s )}{g ( t )} = 21 = 3^{1} \times 7^{1}$
$We recognise that we can apply Prime Matching to match the exponents of corresponding primes.$	$a_{s} - a_{t} = 1$	Theory & Tactics Method Card ALG7 Click to view full details
$We match the exponent for the base 7.$	$b_{s} - b_{t} = 1$
$We match the exponent for the base 11.$	$c_{s} - c_{t} = 0$
$We express s - t in terms of the digit differences.$	$s - t = 100 \times (a_{s} - a_{t}) + 10 \times (b_{s} - b_{t}) + (c_{s} - c_{t})$
$We substitute the differences and calculate the result.$	$s - t = 100 \times 1 + 10 \times 1 + 0 = 110$

Scroll horizontally to view all columns

C