Question

Find 3 digit nos the sum of whose digits is equal to the product

Answer 1

Answer:

NUMBER IS 123 AS:-

1} 1 + 2 + 3 = 6

2}1 x 2 x 3 = 6

Answer 2

Answer: 321 and 123

Step-by-step explanation:

So firSo, we need find digits a,b,c such that a+b+c=abc

Clearly, abc≠0, if a=0,b+c=0⟹b=c=0

So,

a+bab−1=c

which is integer.

(1)If a=3m+1,b=3n+1 or a=3m−1,b=3n−1

3∣(ab−1), but 3∤(a+b)⟹(ab−1)∤(a+b)

(2)If a=3m≤9,b=3n≤9⟹1≤m,n≤3

a+bab−1=3(m+n)9mn−1

⟹(9mn−1)∣(m+n) as (9mn−1,3)=1

But (m+n)≤3+3=6

⟹(9mn−1)≤6 which is clearly impossible as m,n≥1.

(3)If a=3m+1≤9,b=3n−1≤9⟹0≤m≤2,1≤n≤3,

a+bab−1=3(m+n)(3m+1)(3n−1)−1

⟹(3m+1)(3n−1)−1∣(m+n) as ((3m+1)(3n−1)−1,3)=(3(3mn−m+n)−2,3)=1

(3m+1)(3n−1)−1≤5⟹(3m+1)(3n−1)≤6⟹m≤1 and n≤2

If m=1⟹3m+1=4⟹3n−1≤1⟹n<1, but 1≤n≤3

If m=0,

3(m+n)(3m+1)(3n−1)−1 becomes3n3n−2

⟹(3n−2)∣3n⟹(3n−2)∣n as (3n−2,3)=1

So, 3n−2≤n⟹n≤1 ,but 1≤n≤3 so, n=1

So, a=1,b=2⟹c=a+bab−1=3

As a+b+c=abc is symmetric a=3,b=1 and a=3,b=2 will also be solutions corresponding to cases (4) a=3m,b=3n+1 and (5) a=3m,b=3n−1 respectively.

So, the only solution is 1,2,3.

Clearly, 321 is the largest and 123 is the smallest.