Find-the-number-of-integer-solutions-for-a-b-c-d-18900-with-a-b-c-d-1-

Question Number 57985 by mr W last updated on 15/Apr/19

F i n d t h e n u m b e r o f i n t e g e r s o l u t i o n s

f o r a \times b \times c \times d = 18900

w i t h a, b, c, d ⩾ 1 .

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19

18900 = 3 \times 3 \times 3 \times 7 \times 2 \times 2 \times 5 \times 5

18900 = 2^{2} \times 3^{3} \times 5^{2} \times 7^{1}

n u m b e r o f d i v i s o r o f 18900 i n c l u d i n g 1 a n d 18900

a r e (2 + 1) (3 + 1) (2 + 1) (1 + 1) = 72

n o w e x c l u d i n g 1 a n d 18900 \to 72 - 2 = 70

f o r e x a m p l e \dots

a = 2^{2} = 4

b = 3^{3} = 27

c = 5^{2} = 25

d = 7^{1} = 7

4 \times 27 \times 25 \times 7 = 18900

n o w i f a = 2

b = 2 \times 3^{3} = 54

c = 5^{2} = 25

d = 7

2 \times 54 \times 25 \times 7 = 18900

i n t h i s w a y t o t a l i n t e g e r s o l u t i o n

a r e = 70

s i r p l s c h e c k a m i c o r r e c t \dots

Commented by mr W last updated on 15/Apr/19

i {d o n}^{'} t h a v e a n a n s w e r y e t s i r . b u t

i t h i n k t h e r e m u s t b e m u c h m o r e

s o l u t i o n s . e . g .

1 \times 1 \times 1 \times 18900

1 \times 1 \times 18900 \times 1

1 \times 18900 \times 1 \times 1

18900 \times 1 \times 1 \times 1

a r e 4 d i f f e r e n t v a l i d s o l u t i o n s .

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Apr/19

y e s s i r \dots i h a v e e x c l u d e d 1 i n m y s o l u t i o n \dots

Answered by MJS last updated on 17/Apr/19

there are 382 solutions with a ⩽ b ⩽ c ⩽ d but

without this restriction there are exactly

8000 (I hope I {didn}^{'} t count wrong \dots)

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