Question

In the 2001 the United States hosted the
international mathematical olympiad. Let I, M
and O be distinct positive integers such that
the product I.M.O. = 2001, what is the largest
possible values the sum I + M + O?
(B) 55
(C) 99
(A) 23
(DY671​

Answer 1

Answer:

55

Step-by-step explanation:

Try prime numbers that divide 2001

Divisibility by 3 applicable as sum of all digits is divisible by 3

2001/3 = 667

Try other primes..you will see

23 and 29 divided 667

23+29+3 = 55

Answer 2

The largest possible values the sum I + M + O = 671

Given :

In the 2001 the United States hosted the international mathematical olympiad. Let I, M and O be distinct positive integers such that the product I.M.O. = 2001

To find :

The largest possible values the sum I + M + O

(A) 23

(B) 55

(C) 99

(D) 671

Solution :

Step 1 of 2 :

Express 2001 as product of three distinct positive integers

On prime factorisation of 2001 we get

2001 = 3 × 23 × 29

Thus we get

2001 = 1 × 23 × 87 ; We take l = 1 , M = 23 , O = 87

2001 = 1 × 29 × 69 ; We take l = 1 , M = 29 , O = 69

2001 = 1 × 3 × 667 ; We take l = 1 , M = 3 , O = 667

Step 2 of 2 :

Find the largest possible values the sum I + M + O

Now three cases arises :

2001 = 1 × 23 × 87 ; We take l = 1 , M = 23 , O = 87

I + M + O = 1 + 23 + 87 = 111

2001 = 1 × 29 × 69 ; We take l = 1 , M = 29 , O = 69

I + M + O = 1 + 29 + 69 = 99

2001 = 1 × 3 × 667 ; We take l = 1 , M = 3 , O = 667

I + M + O = 1 + 3 + 667 = 671

∴ The largest possible values the sum I + M + O = 671

Hence the correct option is (D) 671

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