## Interview Questions

 Puzzles
 Greengrocer C. Carrot wants to expose his oranges neatly for sale. Doing this he discovers that one orange is left over when he places them in groups of three. The same happens if he tries to place them in groups of 5, 7, or 9 oranges. Only when he makes groups of 11 oranges, it fits exactly. How many oranges does the greengrocer have at least? Ans:  Assume the number of oranges is A. Then A-1 is divisible by 3, 5, 7 and 9. So, A-1 is a multiple of 5×7×9 = 315 (note: 9 is also a multiple of 3, so 3 must not be included!). We are looking for a value of N for which holds that 315×N + 1 is divisible by 11. After some trying it turns out that N = 3. This means that the greengrocer has 946 oranges.
 A cyclist drove one kilometer, with the wind in his back, in three minutes and drove the same way back, against the wind in four minutes. If we assume that the cyclist always puts constant force on the pedals, how much time would it take him to drive one kilometer without wind? Ans:  The cyclist drives one kilometer in three minutes with the wind in his back, so in four minutes he drives 1 1/3 kilometer. Against the wind, he drives 1 kilometer in four minutes. If the wind helps the cyclist during four minutes and hinders the cyclist during another four minutes, then - in these eight minutes. , the cyclist drives 2 1/3 kilometers. Without wind, he would also drive 2 1/3 kilometers in eight minutes and his average speed would then be 17.5 kilometers per hour. So it will take him 3 3/7 minutes to drive one kilometer.
 Two friends, Alex and Bob, go to a bookshop, together with their sons Peter and Tim. All four of them buy some books; each book costs a whole amount in shillings. When they leave the bookshop, they notice that both fathers have spent 21 shillings more than their respective sons. Moreover, each of them paid per book the same amount of shillings as books that he bought. The difference between the number of books of Alex and Peter is five. Who is the father of Tim? Ans:  For each father-son couple holds: the father bought x books of x shillings, the son bought y books of y shillings. The difference between their expenses is 21 shillings, thus x2 - y2 = 21. Since x and y are whole numbers (each book costs a whole amount of shillings), there are two possible solutions: (x=5, y=2) or (x=11, y=10). , Because the difference between Alex and Peter is 5 books, this means that father Alex bought 5 books and son Peter 10. This means that the other son, Tim, bought 2 books, and that his father is Alex.
 On the market of Covent Garden, mrs. Smith and mrs. Jones sell apples. Mrs. Jones sells her apples for two per shilling. The apples of Mrs. Smith are a bit smaller; she sells hers for three per shilling. At a certain moment, when both ladies both have the same amount of apples left, Mrs. Smith is being called away. She asks her neighbour to take care of her goods. To make everything not too complicated, Mrs. Jones simply puts all apples to one big pile, and starts selling them for two shilling per five apples. When Mrs. Smith returns the next day, all apples have been sold. But when they start dividing the money, there appears to be a shortage of seven shilling. Supposing they divide the amount equally, how much does mrs. Jones lose with this deal? Ans:  The big pile of apples contains the same amount of large apples of half a shilling each (from mrs. Jones), as smaller apples of one third shilling each (from mrs. Smith). The average price is therefore (1/2 + 1/3)/2 = 5/12 shilling. But the apples were sold for 2/5 shilling each (5 apples for 2 shilling). Or: 25/60 and 24/60 shilling respectively. , his means that per sold apple there is a shortage of 1/60 shilling. The total shortage is 7 shilling, so the ladies together started out with 420 apples. These are worth 2/5 × 420 = 168 shilling, or with equal division, 84 shilling for each. If Mrs. Jones would have sold her apples herself, she would have received 105 shilling. Conclusion: Mrs. Jones loses 21 shilling in this deal.
 Postman Pat delivers the mail in the small village Tenhouses. This village, as you already suspected, has only one street with exactly ten houses, numbered from 1 up to and including 10. In a certain week, Pat did not deliver any mail at two houses in the village; at the other houses he delivered mail three times each. Each working day he delivered mail at exactly four houses. The sums of the house numbers where he delivered mail were: on Monday: 18 on Tuesday: 12 on Wednesday: 23 on Thursday: 19 on Friday: 32 op Saturday: 25 on Sunday: he never works Which two houses didn't get any mail that week? Ans:  If postman Pat would have delivered mail three times at each house, then the total sum of the house numbers per day would be (1+2+3+4+5+6+7+8+9+10)×3=165. Now that sum is 18+12+23+19+32+25=129. The difference is 165-129=36; divided by 3 this is 12. The sum of the house numbers where no mail was delivered is therefore 12.The following combinations are possible: 2+10, 3+9, 4+8, 5+7, , Each day at four houses the mail was delivered. On Tuesday the sum was 12. 12 can only be made from four house numbers in 2 ways: 1+2+3+6, 1+2+4+5 The same holds for Friday with the sum of 32: 5+8+9+10 6+7+9+10 , N.B.: there are various possibilities for the actual post delivery of the whole week. For example: Monday houses 1, 3, 5 and 9Tuesday houses 1, 2, 3 and 6Wednesday houses 1, 5, 7 and 10Thursday houses 2, 3, 5 and 9Friday houses 6, 7, 9 and 10Saturday houses 2, 6, 7 and 10.
 From a book, a number of pages are missing. The sum of the page numbers of these pages is 9808. Which pages are missing? Ans:  Let the number of missing pages be n and the first missing page p+1. Then the pages p+1 up to and including p+n are missing, and n times the average of the numbers of the missing pages must be equal to 9808: n×( ((p+1)+(p+n))/2 ) = 9808 , In other words: n×(2×p+n+1)/2 = 2×2×2×2×613So: n×(2×p+n+1) = 2×2×2×2×2×613One of the two terms n and 2×p+n+1 must be even, and the other one must be odd. Moreover, the term n must be smaller than the term 2×p+n+1. It follows that there are only two solutions: n=1 and 2×p+n+1=2×2×2×2×2×613, so n=1 and p=9808, so only page 9808 is missing. , n=2×2×2×2×2 and 2×p+n+1=613, so n=32 and p=290, so the pages 291 up to and including 322 are missing. Because it is asked which pages (plural) are missing, the solution is: the pages 291 up to and including 322 are missing.
 Below is an equation that isn't correct yet. By adding a number of plus signs and minus signs between the ciphers on the left side (without changes the order of the ciphers), the equation can be made correct. 123456789 = 100 How many different ways are there to make the equation correct? Ans:  There are 11 different ways: 123+45-67+8-9=100 123+4-5+67-89=100 123-45-67+89=100 123-4-5-6-7+8-9=100 12+3+4+5-6-7+89=10012+3-4+5+67+8+9=100 12-3-4+5-6+7+89=1001+23-4+56+7+8+9=100 1+23-4+5+6+78-9=1001+2+34-5+67-8+9=100 1+2+3-4+5+6+78+9=100