Download Solutions to Exercises on Page 39 #1. All prime numbers are odd

Solutions to Exercises on Page 39
#1. All prime numbers are odd except 2. 3535 is odd. If it can be written as the
sum of two prime numbers, one of the prime numbers must be 2 because sum of
two odd numbers is even. So the sum must be 3535=2+3533. The larger of the
two prime numbers is 3533. Answer: 3533
#2. Since the product of the digits is 2016, we will factor 2016 as product of six
digits and the digits are as large as possible.
2016  2000  16  8  250  8  2  8  252  8  9  28  9  8  7  4 11
Answer: 987411
#3. Factor 1008 as product of six digits and the digits are as large as possible.
1008  9  999  9 112  9  8 14  9  8  7  2 11
The largest six-digit integer is 987211; the smallest six-digit integer is 112789.
The positive difference is 987211-112789=874422. Answer: 874422.
#4. 2520  252 10  9  28  2  5  9  4  7  2  5  23  32  5  7 . Of course you
do not need to completely factor 2520 in order to see that the prime factors of
2520 are 2, 3 5 and 7. 2+3+5+7=17. Answer: 17.
#5. 2250  2  32  53 . Odd prime factors of 2250 are prime factors of 32  53 .
The number of prime factors of 32  53 is 2  1  3  1  3  4  12 . Answer: 12.
#6. 342  9  38  2  32  19 . The number of factors is 1  1  2  1  1  1  12 .
Answer: 12.
#7. 848  8  106  8  2  53  24  53 . 53 is odd and any factor of 53 multiplied by
2 i ( 1  i  4 ) is an even factor. So the ratio is 1 : 4 
1
1
. Answer:
.
4
4
#8. 201984  4  50496  4  8  6312  4  8  8  789  28  789 . Remember that an
integer is divisible by 4 if the number formed by the last two digits is divisible by
4. An integer is divisible by 8 if the number formed by the last two digits is
divisible by 8. 6312  6400  88  8  800  11  8  789 .
Answer: 8.
#9. A trailing zero represents a factor of 10  2  5 . There are many more factors
of 2 than factors of 5 in 40!. From 1 to 40, there are 8 multiples of 5. One of them
25 has two factors of 5. Therefore there are 9 factors of 5 in 40!. Answer: 9.
#10. 8937500  89375  100  89000  3  125  100  89  8  3  125  100
 90  8  5  53  52  4  18  8  1  5  53  52  4 . The last digit of 18  8  1 is 3 and it
is not a multiple of 5. The maximum power of 5 that divides 8937500 is 56 . The
complete factorization is 8937500  22  56  11 13 . Answer: 6.
#11. Between 101 and 199 there are 97 numbers. We divide 97 by 5 and get a
quotient 19. There are 19 multiples of 5 between 101 and 199. We divide 97 by 7
and get a quotient 13. There are 13 multiples of 7 between 101 and 199. If we
add 19 and 13, we get 32. We over counted those numbers that are both
multiples of 5 and 7 (They must be a multiple of 35). There are only 2 multiples
of 35 between 101 and 199. 32-2=30. Answer: 30.
#12. There are 99 numbers strictly between 100 and 200. We divide 99 by 7 and
get a quotient 14. There are 14 multiples of 7. Half of them (7) are divisible by 2.
Divide 14 by 3 and we get a quotient 4. 4 of the multiples of 7 are also multiples
of 3. If we subtract both 7 and 4 from 14, we get 3. But we subtracted those
multiples of 6 twice. There are two of them (divide 14 by 6 to have a quotient 2).
3+2=5. Answer: 5.
#13. 1234567125  1234567 1000  125  125  1234567  8  1 and the second
factor is not divisible by 5. The number formed by the last three digits of
1235672 is 672  600  72  2  3  4  25  8  9 and it is divisible by 8. 103 is the
maximal power of 10 that divides the product. Answer: 3.
#14. Let x=ABC and y=DEF. The equation becomes 51000 x  y   61000 y  x  .
4994 x  5995 y . 11  454  x  11  545  y . 454  x  545  y . The prime factorization
of 454 is 2  227 and the prime factorization of 545 is 5 109 . They are coprime.
545 divides x. Since x is a three-digit number, x=545 and y=454. The original
six-digit number is 545,454. Answer: 545,454.
#15. There are 3 multiples of 5’s from 1 to 19. The last three digits of 19! are
zeros. 19! is divisible by 9, so the sum of the digits is divisible by 9. The sum of
the digits except the missing one is 1+2+1+6+4+5+1+8+8+3+2=4*9+5. The
missing digit is 4. Answer: 4.
#16. Any three consecutive integers will contain a multiple of 3. Any six
consecutive integers will contain 2 multiples of 3. The product of any six
consecutive integers is a multiple of 9. Since the sum of the digits is a multiple of
9, we see the missing digit is 8. (Actually 11 12 13 14 15 16 17  98017920 .
So the question should be changed to “The product of seven consecutive
integers …”.) Answer: 8.
#17. The mean of three consecutive numbers is the middle one.
13800  138 100  2  69  4  25  2  3  23  4  25  23  24  25 . The three
consecutive integers are 23, 24 and 25. The mean is 24. Answer: 24.
#18. 1000000  106  26  56 . Dividing by 2 then multiplying by 5 is equivalent to
replacing a factor of 2 by 5 if 2 is a factor. Otherwise it will not be divisible by 2
and we will get a fractional number. The last integer will be after 6 steps of
replacing a factor of 2 by 5 and we get 56  56  512 . Answer: 512 .
#19. 255  5  51  5  3 17  15  17 . Answer: 17.
#20. 1320  10  132  10 1112 . Answer: 12.
#21. From 1 to 9, there are 4 multiples of 2, 2 multiples of 4 and 1 multiple of 8.
m  4  2 1  7 .
From 1 to 9, there are 3 multiples of 3 and 1 multiple of 9. n  3  1  4 .
m  n  11 .
Answer: 11.
#22. 2014  2 1007  2  19  53 .
Answer: 53.
#23. Using the above prime factorization of 2014 we know that there are
1  11  1(1  1)  8
factors of 2014. These 8 factors can be paired together so
that each pair has product 2014. There are 4 pairs. So n=4. Answer: 4.
12!
is a factor of 12! that is
n
12!
12!
12!
 12 .
less than
can be any of the numbers from 1 to 11, i.e. n 
, for
11!
n
i
i=1, 2, …, 11. Answer: 11.
#24. If n is a factor of 12! that is greater than 11! ,
#25. If the three distinct digits selected are a, b and c, the six possible three-digit
numbers formed are abc  100a  10b  c , acb  100a  10c  b ,
bac  100b  10a  c , bca  100b  10c  a , cab  100c  10a  b and
cba  100c  10b  a . Their sum is S  222a  b  c  . We see that 222 is a
common factor of all these sums.
Picking the three digits 1, 2, 3 and then the three digits 1, 2, 4, we see that there
are no other common factors. Answer: 222.
#26. If n is a positive integer such that, when divided by 10, 9, 8, …, 2 gives a
remainder 9, 8, 7, …, 1, then n+1 is divisible by all the numbers from 10, 9, 8, …, 2.
This is equivalent to say that n+1 is a multiple of the least common multiples of 2,
3, 4, .., 10. The least common multiples of these numbers is 23  32  5  7 . (We just
multiply the highest power of these primes from 2 to 10). The smallest such n is
23  32  5  7  1  2  5  4  9  7  1  10  252  1  2520  1  2519 . Answer: 2519.
4 x  13 4 x  2  15 4 x  2
15
22 x  1
15
15





2
. For it
2x  1
2x  1
2x  1 2x  1
2x  1
2x  1
2x  1
to be an integer, 2 x  1 must divide 15. 2 x  1  1 , 2 x  1  3 , 2 x  1  5 ,
2 x  1  15 . This gives 8 distinct integer solutions for x. Answer: 8.
#27.
#28. 2010  20110  3  67  2  5  2  3  5  67 . 67 is a prime number. Any factor of
2010 that is less than 60 cannot be divisible by 67. So it must be a factor of
2  3  5  30 . There are 1  11  11  1  8 such factors. Answer: 8.
#29. The two positive integers differ by 3 and have a product 154. Factor 154 as
the product of two such numbers: 154  2  77  14  11 . 14+11=25. Answer: 25.