18

Assume that initial there were 3*X bullets.

So they got X bullets each after division.

All of them shot 4 bullets. So now they have (X - 4) bullets each.

But it is given that,after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X

Therefore, the equation is

3 * (X - 4) = X

3 * X - 12 = X

2 * X = 12

X = 6

Therefore the total bullets before division is = 3 * X = 18

Find sum of digits of D.

Let

A= 19991999

B = sum of digits of A

C = sum of digits of B

D = sum of digits of C

(HINT : A = B = C = D (mod 9))

Answer

The sum of the digits od D is 1.

Let E = sum of digits of D.

It follows from the hint that A = E (mod 9)

Consider,

A = 19991999

< 20002000

= 22000 * 10002000

= 1024200 * 106000

< 10800 * 106000

= 106800

i.e. A < 106800

i.e. B < 6800 * 9 = 61200

i.e. C < 5 * 9 = 45

i.e. D < 2 * 9 = 18

i.e. E 1).

1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded. 2. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on. 3. On the nth and last day, the remaining n medals were awarded. How many days did the contest last, and how many medals were awarded altogether? Answer

Total 36 medals were awarded and the contest was for 6 days.

On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals On day 6: Medals...

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