Hello, 0110kim!

A cube with edge length 2 metres is cut up into cubes

each with edge lengths 5 centimetres.

If all these cubes were stacked one on top of the other to form a tower,

the height of the tower would be:

. . (a) 32 km . . (b) 160 m . . (c) 1600 m . . (d) 3.2 km . . (e) 320 m

The cube is 2 meters = 200 cm on an edge.

It is cut into 5 cm cubes; each edge is cut into 40 parts.

There are: .40 × 40 × 40 .= .64,000 cubes.

A stack of 64,000 cubes, each 5 cm high, would be: .5 × 64,000 .= .320,000 cm high.

And: .320,000 cm .= .3200 m .= .3.2 km. . . answer (d)

I have a very primitive solution . . .A number of less than 2008; it is odd.

It leaves a remainder of 2 when divided by 3

and a remainder of 4 when divided by 5.

What is the sum of digits of the largest such number?

. . (a) 26 . . (b) 25 . . (c) 24 . . (d) 23 . . (e) 22

Let the number beN.

WhenNis divided by 5, the remainder is 4.

. . ThenNis of the form: .5a + 4

For any integer value ofa,5aends in 0 or 5.

. . So: .N .= .5a + 4 .ends in 4 or 9.

But we are told thatNis odd . . . Hence,Nends in 9.

Since N < 2008, the choices are: .1999. 1989, 1979, 1969, . . .

Which of these has a remainder of 2 when divided by 3?

. . 1999 ÷ 3 .= .666, rem. 1

. . 1989 ÷ 3 .= .663, rem. 0

. . 1979 ÷ 3 .= .659, rem. 2 . ← . There!

Therefore: .N = 1979, and the sum of the digits is26. . . answer (a)