Mathemagics 5: Solving two interesting problems

Have a look at these two interesting problems, which will make you love maths.

GOOD NUMBER

A positive integer is called a “good number” if it is equal to four times of the sum of its digits.

Now a number with four or more digits cannot be good number, as for the greatest number :

4(9+9+9+9) = 144 which is not a four digit number.

Now for a three digit number:

Ø  100a+10b+c = 4(a+b+c)

Ø  96a+6b = 3c

If b = 0, 32a = c, which is not possible.

Hence good number is obviously some two digit number.

Ø  10a+b = 4(a+b)

Ø  6a = 3b

Ø  2a = b

Hence there can be four possibilities 12, 24, 36 and 48.

 

A three-digit numbers such that each is equal to the sum of the factorials of its own digits.

Let abc = 100a+10b+c be desired three digit number 7! = 5040 indicates that a, b, c ≤ 6, and further, if one of a,b,c is 6 then Þ = abc > 6! = 720 that means one digit should be greater than 6 but that’s impossible, So a,b,c  £ 5  . Since  555 > 5! + 5! + 5! . So a,b,c cannot be all 5  and the hundredth place digit cannot be 5. On the other hand, 4! + 4! + 4! = 72 which is not a three digit number, So at least one of a,b,c is 5.

For the first digit we have three choices 1,2,3

If a = 1, b = 5 then c = 0,1,2,3,4. When a = 1, c = 5 then b = 0,1,2,3,4.

If a = 2 only possibility is b,c = 5

If a = 3 only possibility is b,c = 5.

Checking all these combinations, we get 145 as the unique number which is equal to the sum of the factorials of its own digits.