Solve the following equation if N, O & G are single digit numbers and NO & GN are double digit numbers. Find the values for N, O & G.
learn_more caption=”Solution explanation”]
- N can only be equal to either 1 or 2; otherwise G is a two digit number. (i.e., N + N + N + N = G or N x 4 = G)
- N has to be an even number (i.e., O + O + O + O = N or O x 4 = N)! Therefore, N = 2, since 4 times any number will always result in an even number.
- Since O is still unknown, G (G = N x 4) can only be equal to either 8 or 9; 8 if there is no carry over from the Equation O x 4 = N (where N x 4 = G or = 8), or 9 if there is a carry over from the Equation O x 4 = N (where (N x 4) +1 = G or = 9).
- Solving for O without a carry over:
- ((Nx10) + O) x 4 = 82
- (20 + O) x 4 = 82
- 80 + (4 x O) = 82
- 4 x O = 2
- O = 0.5 (not valid, must be a whole number)
- Solving for O with a carry over:
- [((Nx10) + O) x 4] = 92
- (20 + O) x 4 = 92
- 80 + (4 x O) = 92
- 4 x O = 12
- O = 3 (valid answer)
- Now check the solution:
- NO + NO + NO + NO = GN
- 23 + 23 + 23 + 23 = 92
A ball travels a certain distant before bouncing, half as far before the next bounce, and so on. If it has traveled 93 meters at the fifth bounce, how far was the first bounce?
We know the ball bounces five times in 93 meters. We also know that the ball bounces half the distance each time. Working backwards, we can write this as:
93 = 1x + 2x + 4x + 8x + 16x
Therefore x = 3 and the first bounce was 16 x 3 = 48 meters
With Gear #1 stationary, how many complete turns does the right gear (#2) make to go around the left gear (#1) once?
Gear #2 turns twice.
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