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Digital and Skeleton Puzzles Mental Calculation Set 2

This is a timed quiz. You will be given 400 seconds to answer all questions. Are you ready?


Can the reader give the sum of all the whole numbers that can be formed with the four figures I, 2, 3, 4? That is, the addition of all such numbers as 1,234,1,423,4,312, etc. You can, of course, write them all out and make the addition, but the interest lies in finding a very simple rule for the sum of all the numbers that can be made with four different digits selected in every possible way, but zero excluded.

Correct! Wrong!

If you multiply 6,666 by the sum of the four given digits you will get the correct answer. As I, 2, 3, 4 sum to 10, then 6,666 multiplied by 10 gives us 66,660 as our answer. Taking all possible selections of four different digits, the answer is 16,798,320, or 6,666 X 2,520.

Professor Rackbrane wants to know what is the sum of all the numbers that can be formed with the complete nine digits (0 excluded), using each digit once, and once only, in every number?

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There are several ways of attacking this puzzle, and the answer is 201,599,999,798,400. The sum of the digits is 45 and 45 X 8! = 1,814,400 Now write down 18144 18144 18144 18144 to nine places, add up and put 00 at the end, and there is the answer

What is the smallest number composed only of the digits 3 and 7 that may be divided by 3 and by 7, and also the sum of its digits by 3 and by 7, without any remainder? Thus, for example, 7733733 is divisible by 3 and by 7 without remainder, but the sum of its digits (33), while divisible by 3, is not divisible by 7 without remainder. Therefore it is not a solution.

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The smallest number possible is 3,333,377,733, which is divisible by 3 and by 7, and the sum of its digits (42) also divisible by 3 and by 7. There must be at fewest three 7's and seven 3's, and the 7's must be placed as far to the right as possible. 120

If you arrange the nine digits," said Professor Rackbrane, "in three numbers thus, 147, 258, 369, they have a common difference of III, and are, therefore, in arithmetical progression." Can you find four ways of rearranging the nine digits so that in each case the three numbers shall have a common difference and the middle number be in every case the same?

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If the nine digits are written at haphazard in any order, for example, 4 1 2539768, what are the chances that the number that happens to be produced will be divisible by 37 without remainder

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Call the required numbers ABCABCABC. If the sum of the A digits, the B digits, and the C digits respectively are: 1, 2, 3 None

What number composed of nine figures, if multiplied by 1,2, 3,4, 5, 6, 7, 8, 9, will give a product with 9, 8, 7, 6, 5, 4, 3, 2, 1 (in that order), in the last nine places to the right?

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989,010,989 multiplied by 123,456,789 produces 122,100,120,987,654,321, where the last nine digits are in the reverse order.

The professor, a few mornings ago, proposed that they should find all those numbers composed of three different digits such that each is divisible without remainder by the square of the sum of those digits. Thus, in the case of 112, the digits sum to 4, the square of which is 16, and 112 can be divided by 16 without remainder, but unfortunately 112 does not contain three different digits. Can the reader find all the possible answers?

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The numbers are 162,243,324,392,405,512,605,648,810, and 972. These, we think, are all the cases that exist.

The following is a rather curious puzzle. Find the smallest number that, when divided successively by 45, 454, 4545 and 45454, leaves the remainders 4, 45, 454, and 4,545 respectively. This is perhaps not easy but it affords a good arithmetical exercise.

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The smallest number that fulfills the conditions is 35,641,667,749. Other numbers that will serve may be obtained by adding 46,895,573,610 or any multiple of it.

Professor Rackbrane recently asked his young friends to find all those fivefigure squares such that the number formed by the first two figures added to that formed by the last two figures should equal a cube. Thus with the square of 141, which is 19,881, if we add 19 and 81 together we get 100, which is a square but unfortunately not a cube. How many solutions are there altogether?

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There are three solutions. They are 56,169 (the square of 237), where 56 + 69 = 125 (thecubeof5); 63,001 (the square of251), where 63 + 01 = 64 (the cube of 4); 23, I 04 (the square of 152), where 23 + 04 = 27 (the cube of3).

Arrange the ten digits, 1 2 3 4 5 678 9 0, in such order that they shall form a number that may be divided by every number from 2 to 18 without in any case a remainder. As an example, if I arrange them thus, 1 2 7 4 9 5 3 6 8 0, this number can be divided by 2, 3,4, 5, and so on up to 16, without any remainder, but it breaks down at 17

Correct! Wrong!

There are four solutions, as follows: 2,438,195,760; 3,785,942,160; 4,753,869,120; 4,876,391,520. The last figure must be zero. Any arrangement with an even figure next to the zero will be divisible by 2, 3, 4, 5, 6, 9, 10, 12, 15, and 18. We have therefore only to consider 7, 11, 13, 16, and 17. To be divisible by 11 the odd digits must sum to 28 and the even to 17, or vice versa. To be divisible by 7 X 11 X 13 = 1001, if we ignore the zero, the numbers formed by the first three and the last three digits must sum to the middle three. (Note that the third case above is really 474-1386-912, with the 1 carried forward and added to the 4.) But, we cannot do better than take the lowest multiple (82) of the lowest common multiple of the divisors (12,252,240), which gives ten figures (this is 1,004,683,680), and keep on adding that lowest common multiple until all digits are different. The 199th multiple will give us the first answer, 309 the second, 388 the third, and 398 the fourth. The work can be considerably shortened by leaping over groups where figures will obviously be repeated, and all the answers may be obtained in about twenty minutes by the use of a calculating machine.

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