Find-the-sum-of-all-positive-two-digit-integers-that-are-divisible-by-each-of-their-digits-

Question Number 110591 by Aina Samuel Temidayo last updated on 29/Aug/20

Find the sum of all positive

two - digit integers that are divisible

by each of their digits .

Answered by Rasheed.Sindhi last updated on 30/Aug/20

L e t N = 10 t + u i s a n u m b e r s u c h

t h a t u ∣ 10 t + u \land t ∣ 10 t + u

10 t + u = m u \land 10 t + u = n t

(N i s c o m m o n m u l t i p l e o f m & n)

u = \frac{10 t}{m - 1} \land u = t (n - 10)

\frac{10 t}{m - 1} = t (n - 10)

t \neq 0 (∵ T h e n u m b e r i s o f 2 - d i g i t)

\frac{10}{m - 1} = n - 10

(m - 1) (n - 10) = 10

m - 1 = a \land n - 10 = 10 / a

[W h e r e a ∣ 10]

(m, n) = (a + 1, 10 + \frac{10}{a})

A s N = c o m m o n m u l t i p l e o f m & n

o r a + 1 & 10 + \frac{10}{a} .

S o,

N = k lcm (a + 1, 10 + \frac{10}{a}) for some k

(\in N)

F o u r p o s s i b l e c a s e s

^{★ 1} m - 1 = 1 \land n - 10 = 10

(m, n) = (2, 20)

N = k lcm (2, 20)

N = \overset{\times}{20}, \overset{\times}{40}, \overset{\times}{60}, \overset{\times}{80} (N o v a l u e o f N)

^{★ 2} m - 1 = 2 \land n - 10 = 5

(m, n) = (3, 15)

N = k lcm (3, 15)

N = 15, \overset{\times}{30}, \overset{\times}{45}, \overset{\times}{60}, \overset{\times}{75}, \overset{\times}{90}

^{★ 3} m - 1 = 5 \land n - 10 = 2

(m, n) = (6, 12)

N = k l c m (12, 6)

N = 12, 24, 36, 48, \overset{\times}{60}, \overset{\times}{72}, \overset{\times}{84}, \overset{\times}{96}

^{★ 4} m - 1 = 10 \land n - 10 = 1

(m, n) = (11, 11)

N = k lcm (11, 11)

N = 11, 22, 33, 44, 55, 66, 77, 88, 99

R e d n u m b e r s a r e s u c c e s s f u l .

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

Thanks but please try to complete it . I

post questions because I {don}^{'} t really have any idea

of how to solve them .

Commented by Rasheed.Sindhi last updated on 30/Aug/20

C o m p l e t e d t h e a n s w e r

Answered by floor(10²Eta[1]) last updated on 29/Aug/20

first of all, we have 3 obvious things :

(1) : this works with all multiple of 11

(2) : this {doesn}^{'} t work with any prime

number (except 11)

(3) : this {doesn}^{'} t work with any multiple

of 10 (because we {can}^{'} t say that 0 divides

something)

10 a + b s . t . a ∣ 10 a + b \land b ∣ 10 a + b

\Rightarrow a ∣ b and b ∣ 10 a

so or b ∣ 10 \Rightarrow b = 1, 2 or 5

or b ∣ a \Rightarrow a = b (because a ∣ b)

but when a = b is the 1 case that i said

so {let}^{'} s see when b = 1, 2, 5

b = 1 \Rightarrow (11, 21, 31, 41, 51, 61, 71, 81, 91)

works : (11)

b = 2 \Rightarrow (12, 22, 32, 42, 52, 62, 72, 82, 92)

works : (12, 22)

b = 5 \Rightarrow (15, 25, 35, 45, 55, 65, 75, 85, 95)

works : (15, 55)

only cases that works :

(12, 15, and all multiples of 11)

so the sum of these numbers are

12 + 15 + 11 (1 + 2 + 3 + 4 + \dots + 9)

= 27 + 11.45 = 522

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

Thanks but I {don}^{'} t think this is

correct .

Commented by floor(10²Eta[1]) last updated on 30/Aug/20

? ? ? ? ? why

Answered by mr W last updated on 30/Aug/20

n u m b e r s i n f o r m a a a r e c l e a r, o t h e r

n u m b e r s m u s t b e c h e c k e d o n e b y o n e .

w e g e t t o t a l l y 14 s u c h n u m b e r s :

11

12

15

22

24

33

36

44

48

55

66

77

88

99

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