# Computation Brain Training the power of numbers.

WELCOME TO PUZZLES-ONLINE-NICHE

Below you will find several varieties of computation puzzles- including Computation Cross,

Computation Square, and Computation word math puzzles.

COMPUTATION CROSS 1-4

Rules:

These computation brain training puzzles are completed just like regular crosswords, only

letters are replaced by numbers, and the numbers are entered one number per square

The clues are the sum of the boxes given.

For example the first clue(1- across) for CROSS1 , is 13.

Looking at the diagram we see there are two squares for 1-across.

So their sum of those two squares must equal 13.

For computation cross1 the answer for clue 1-across is 6 in square one, followed by 7 in

square two, (because 6+7 =13).

Computation cross1

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Clues Computation cross1

Clues:

 ACROSS DOWN 1. 13 2. 16 3. 13 3. 15 5. 22 4. 5 8. 8 6. 16 10. 17 7. 15 11. 15 9. 13 12. 11 10. 20 13. 12 12. 9 15. 17 14. 8 17. 13 15. 14 18. 12 16. 16

Computation cross2

 1 2 7 3 4 5 6 7 8 9 6 10 7 11 8 12 13 14 15 16 17 18 7

Clues:

 ACROSS DOWN 1. 34 1. 28 5. 17 2. 21 6. 11 3. 10 8. 30 4. 17 10. 16 5. 32 11. 15 7. 31 12. 28 9. 20 14. 24 11. 28 17. 20 13. 24 18. 33 15. 11 16. 13

Computation cross3

 1 2 3 4 5 6 7 8 9 10 7 11 12 4 6 13 6 14 15 16 17 18 19

Clues:

 ACROSS DOWN 1. 5 2. 12 3. 5 3. 9 6. 23 5. 13 7. 35 7. 15 8. 15 9. 13 10. 17 10. 20 13. 13 12. 9 15. 10 14. 8 16. 12 15. 14 17. 14 16. 16

Computation cross4

 1 2 7 3 4 5 6 7 8 9 10 7 11 12 8 13 14 15 16 17 18 19 20 6

Clues:

 ACROSS DOWN 1. 35 1. 20 5. 15 2. 20 6. 20 3. 17 8. .30 4. 24 10. 12 5. 34 11. 20 7. 32 13. 13 9. 18 15. 24 12. 23 17. 22 14. 18 19. 20 16. 15 20. 30 18. 16

COMPUTATION SQUARE 1-2

Rules:

These puzzles consist of a five x five board ( 5 x5).

Only the numbers 1-9 are used, and only one digit per square is allowed.

A digit may appear more than once in a solution.

A prime number is one that is divisible by ONLY the number 1 and itself ( 3, 7, 11, are prime).

Computation Square1

 1 2 3 4 5 6 7 8 9 10

 Clues:ACROSS1. The square of a square ( odd)3. Consecutive digits5. Consecutive digits ( not in order)7. Even digits ( all different)8. Twelve times a prime number.9. Product of two primes.10. The square of the cube root of 6-down. DOWN 1 A multiple of the square root of 10-across 2. The sum of the first and third digits is equal to the fourth; the second and fifth digits are the same. 3. A palindrome of even numbers. 4. The cube of a cube (even) 6. A palindrome that is the cube of a prime number. 8. A trombone number

Computation Square2

 1 2 3 4 5 6 7 8 9

 Clues:ACROSS1. The square of thethird smallest primenumber.3. The square of an Arabic number which looks like a 24. An odd number that is thirty less than it would be if written upside down.6. Onion Market Day(Bern).8. The square of a prime number larger than 1-across, but smaller than 3-across.9.The next square after 8-across. DOWN1The sum of the last two digits equals the sum of the first three.2. The second and fourth digits are alike.3.The square of an even number which is itself a square.5. The next square after 3-across.7. The sum of its digits is the square root of 9-across.

Computation Brain Training: WORD MATH PUZZLES

MAY DAY

Thirteen children from Miss Kendrick's homeroom class have been selected to participate

in this year's May Day festivities.

They must choose a class leader to captain the project.

Miss Kendricks has the children form a circle and count clockwise, with each count of

thirteen being reached that child will be eliminated from the game, and the count restarts to the

left of the last eliminated child.

Little Cassie Wilson volunteers to begin the count. Because she volunteered Miss Kendricks

allows Cassie to pick which child the count begins with.

Call Cassie child A, and the rest of the children in the circle B-M.

With which child should Cassie choose the count to begin to ensure she is the last one

left, and hence wins the game?

THE MOPED INCIDENT:

Jack had to go to the city - 15 miles away from home. So he hopped on his trusty moped ( at a top

speed of 20mph), and zoomed into town. Jack picked up his fandangle and prepared for the

return trip leaving the city at 2pm. But after riding only two-thirds of the way home, Jack's moped

got a flat tire and he was forced to walk the rest of the way.

Jack arrived home at 330pm, and was immediately assailed by his wife who

demanded to know what took him so long. When he explained about the moped incident,

she only looked at him with consternation and said he should have walked faster.

Exactly how fast did Jack walk?

MOPED INCIDENT PART DEUX

A few weeks later Jack and his wife went on a holiday trip using their Chrysler station wagon,

as the wife refused to be seen on the back of Jack's moped.

The only thing his spouse was grateful for was that Jack always adhered to the posted speed limits

on both local and state highways.

Assuming that the posted limits, respectively were 35mph and 55mph, and their trip took 10 hours

(in which they traveled 470 miles), How much of their travel was on local highways

and how much on interstates?

Cassie should pick child G to start.

Then the order of elimination will be: F, G, I, L, C, K, H, J, B, D, E, M.

The key is to use the formula which gives the relationship between speed, distance, and time.

Speed is given as the distance traveled over time. Mathematically we write this using symbols :

Speed (s) = distance (d) over (divided by /) time (t), or s = d / t.

Now lets apply this to our particular situation. We are given s =20mph, we know the initial time

he left was 2pm, but we need to calculate the distance(d).

We are told Jack traveled 2/3 of the way home, and we know the total distance was 15 miles

(from home to city), thus Jack traveled (15 divided by 2/3) or 10 miles , when the tire gave out.

Thus it took Jack how long to travel 10 miles? Well if he was traveling at 20mph, which means

in one hour he would have gone 20miles, then Jack must have taken 1/2 hour to travel 10 miles.

Which you can verify by substituting t = d / s, or 10mile/20mph = 1/2 hour.

Further Jack left at 2pm, therefore traveled 1/2 hour and blew his tire.

He must have blown his tire at 2:30pm.

Jack now walks the rest of the way ( which is 15 miles - 10 miles) = 5 miles.

We now know he walked from 2:30 pm until he arrived home at 3:30pm, therefore he walked 5 miles

in one hour. ( When we ask for rate, we really mean speed), and we know speed =

d / t, or substituting for the variables , we have Speed(s) = 5miles / 1 hour = 5 mph.

Jack walked home at the rate of 5 miles per hour.

Again, we make use of the formula s = d / t.

We are given t = 10 hours, and d = 470 miles.

Therefore we have s = 47 Mph.

Since this is well over 35mph, but under 55 mph, we can conclude they probably spent more

time on the interstate, than the local roads.

The difference in speeds was 20mph, and the average of 47mph, was 8 under the 55mph,

but 12 over the 35 mph speed limit. Using this ratio (8/12) we can figure how much time was

spent driving at the 35mph limit and the 55 mph limit.

Because we have 8/12 or 4/6 , we can see that they spent 4 hours driving at 35mph and 6 hours

driving at 55mph, and then we can plug in our formula as a check.

we have d = s x t, or

d1= 35 x 4 hours = 140 miles.

d2 = 55 x 6 hours = 330 miles.