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Question Number 163168 by kdaramaths last updated on 04/Jan/22

Commented by mr W last updated on 04/Jan/22

$q u e s t i o n i s w r o n g!$ $t h e r e i s n o s o l u t i o n! s e e m y p r o o f$ $b e l o w .$
	Answered by mahdipoor last updated on 04/Jan/22

	$d i v i s o r s o f n =$ $D_{0}) 1, 5, 5^{2}, . . ., 5^{α}$ $D_{1}) 119, 119 \times 5, 119 \times 5^{2}, . . ., 119 \times 5^{α}$ $. . .$ $D_{β}) 119^{β}, 119^{β} \times 5, . . ., 119^{β} \times 5^{α}$ $\Rightarrow$ $Σ D_{i} = 119^{i} (1 + 5 + 25 + . . . + 5^{α}) = \frac{119^{i} (5^{α + 1} - 1)}{4}$ $S u m o f d i v i s o r s o f n = Σ D_{0} + Σ D_{1} + . . . + Σ D_{β} =$ $\frac{5^{α + 1} - 1}{4} (1 + 119 + 119^{2} + . . . + 119^{β}) =$ $\frac{(5^{α + 1} - 1) (119^{β + 1} - 1)}{4 \times 118} = 3720$ $\Rightarrow (5^{α + 1} - 1) (119^{β + 1} - 1) = 1755840$ $= 124 \times 14160$ $\Rightarrow α = 2 β = 1$
	Commented by mr W last updated on 04/Jan/22

	$h a v e y o u c h e c k e d t h e s u m o f d i v i s o r s$ $o f t h e n u m b e r 5^{2} 119 (= 2975) ?$ $i t i s 4464, n o t 3720!$
	Commented by kdaramaths last updated on 04/Jan/22
	thanks
	Commented by mr W last updated on 04/Jan/22

	Commented by mahdipoor last updated on 04/Jan/22

	$i t h i n k 119 i s p r i m e n a m b e r!, y o u a r e$ $r i g h t M r w, m y a n s i s w r o n g .$
	Commented by mr W last updated on 05/Jan/22

	$y o u r m e t h o d i s c o r r e c t .$ $m a y b e t h e p e r s o n w h o w r o t e t h e$ $q u e s t i o n a l s o t h i n k s t h a t 119 i s a$ $p r i m e n u m b e r, b e c a u s e h e e x p e c t e d$ $e x a c t l y t h a t w h a t y o u d i d .$
	Answered by mr W last updated on 04/Jan/22

	$n = 5^{α} 119^{β} = 5^{α} 7^{β} 17^{β}$ $α, β ⩾ 1$ $S = \underset{p = 0}{\sum^{α}} \underset{q = 0}{\sum^{β}} \underset{r = 0}{\sum^{β}} 5^{p} 7^{q} 17^{r}$ $S = \frac{5^{α + 1} - 1}{5 - 1} \times \frac{7^{β + 1} - 1}{7 - 1} \times \frac{17^{β + 1} - 1}{17 - 1}$ $S = \frac{(5^{α + 1} - 1) (7^{β + 1} - 1) (17^{β + 1} - 1)}{384} = 3720$ $s o t h e q u e s t i o n i s t o f i n d α, β \in N,$ $s u c h t h a t$ $(5^{α + 1} - 1) (7^{β + 1} - 1) (17^{β + 1} - 1) = 3720 \times 384$ $5^{α + 1} - 1 < \frac{3720 \times 384}{(7^{2} - 1) (17^{2} - 1)}$ $α < \log_{5} [\frac{3720 \times 384}{(7^{2} - 1) (17^{2} - 1)} + 1] - 1 \approx 1.88$ $\Rightarrow α = 1$ $(7^{β + 1} - 1) (17^{β + 1} - 1) < \frac{3720 \times 384}{(5^{2} - 1)} = 59520$ $\Rightarrow β < 1.31$ $\Rightarrow β = 1$ $b u t w i t h α = 1, β = 1 :$ $S = \frac{(5^{2} - 1) (7^{2} - 1) (17^{2} - 1)}{384} = 864 < 3720$ $w i t h α = 2, β = 1 :$ $S = \frac{(5^{3} - 1) (7^{2} - 1) (17^{2} - 1)}{384} = 4464 > 3720$ $w i t h α = 1, β = 2 :$ $S = \frac{(5^{2} - 1) (7^{3} - 1) (17^{3} - 1)}{384} = 104994 > 3720$ $\Rightarrow t h e r e e x i s t n o α, β s u c h t h a t$ $t h e s u m o f d i v i s o r s o f 5^{α} 119^{β} i s$ $3720!$
	Commented by Tawa11 last updated on 06/Jan/22

	$Great sir$
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