Download Category 3 Number Theory Meet #1 October 2007 – Practice #2

Category 3
Number Theory
Meet #1 October 2007 – Practice #2
1) When I open my mathematics book, there are two pages which face me and the product of
the two pages is 1806. What are the two page numbers?
Since the two pages are consecutive they must be close to the square root of the product.
Since 40x40 = 1600 and 45x45 = 2025, the two numbers must be between 40 and 45. Since
the product ends in 6, logic tells me the two numbers must be 42 and 43 to get a product that
ends in 6. Checking confirms that 42 & 43 are the consecutive numbers with product 1806.
1
then § ( 3 + ↑2↑ ) § = ?
2) If ↑N↑ = N2 + 1 and §N§ =
N
§ (3 + 22 + 1) § = § 8 § = 1/8
3) What is the largest three-digit number that is divisible by both 7 and by 8?
If the number is divisible by 7 and 8, the number must be divisible by 56. By looking at
groups of 56 you can try to get as close to 1000 as possible. Ten 56's make 560, five 56's
would be 280. A total of fifteen 56's would be 840. Another two 56s(112) would bring the
total to 952 and any more 56s would be more than 1000
4) When the six-digit number 3456N7 is divided by 8, the remainder is 5. Find the sum of
all possible values of N.
If the remainder is 5, subtracting 5 would make the number divisible by 8. For 3456N2 to be
divisible 8, 6N2 would have to be divisible by 8. Since 600 is divisible by 8, the last two
digits, ending in 2, must be divisible by 8. The possibilities there would be 32 and 72. So the
possible values of N are 3 & 7.
5) Find the cube of whichever of the following numbers: 1, 2, 7, 8, or 10 can never be the
difference of two prime numbers.
Since all primes other than 2 are odd, they are all an even number apart(except when
comparing to 2). So primes can definitely be 2, 8, or 10 apart(examples (3,5); (3,11)
;(3,13)). So the answer is either 1 or 7. Since 2 + 1 = 3 is prime and 2+7 = 9 is not prime.
No two primes are even 7 apart.
6) The sum of all of the integers from 1 to 30 is 465. What is the sum of all the integers
from 1 to 30 that are divisible by 3?
We need the sum of the numbers 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 which is 165
7) 160,083 = AB x CC x DB (where A>D)
Evaluate : C5 + 5B4 ÷ B3 – AD
160083 = 112 x 33 x 72 So A = 11, B = 2, C = 3, D = 7
C5 + 5B4 ÷ B3 – AD
35 + 5(24) ÷23 – 11(7) = 243 +5(16) ÷8 – 77 = 243 + 10 – 77 = 176
8)
N = the sum of all prime numbers between 60 and 70
W = the mean of all two digit composite numbers whose tens digit is 8
Find the value of N – W expressed as a decimal
N = 61+67 = 128
W = (80+81+82+84+85+86+87+88)/8 = 84.125
N – W = 128 – 84.125= 43.875
9) 2E57E is a five digit number which is divisible by 12. Find the value of E.
In order to be divisible by 12 a number must be divisible by 3 and 4. To be divisible by 4 the
last 2 digits must forma number divisible by 4. In the 70s that gives us the option of 72 or 76.
So E = 2 or 6. The sum of the digits though must be divisible by 3. If E =6, the sum would be
26 which is not divisible by 3. If E =2, the sum of the digits is 18! So E = 2