Hello

from Debbie Cummins. Australia

I am a Mum of a 12 yr.boy needing help with some math problems. I need not only the answers but how it is worked out.

  1. Both the leftmost digit & the rightmost digit of a 4 digit number N are equal to 1. When these digits are removed, the 2 digit number thus obtained is N div by 21 Find N.
  2. Find all 3 digit even numbers N such that 693xN is a perfect square, that is, 693x N = k x k where k is an integer.
  3. The parliament of the land of Achronia consists of 2 houses. The parliament was elected in 1995 for a period of 4 years beginning on Monday, the 1st.January 1996 when the 2 houses had their first sessions. Accordind to the rules, the meetings of the first house must occur every 10 days for the duration of the term, & the meetings of the second house must occur every 12 days. For example, the second meetings of the first and the second houses were held on the 11th. and 13th. of January respectively. A new law can be passes only when both houses meet on a Monday. How many opportunities will the parliment members have to pass new laws during this 4 year term?
  4. When a certain number N is Divided by d, the remainder is 7. If the original number N is multiplied by 5 and then divided by d, the remainder is 10. Find d.
  5. Find the sum of all postive integers not greater than 10000 that are divisible by either 3 or 11 but not by both of them.
  6. 91 five -digit numbers are written on a blackboard. Prove that one can find three numbers on the blackboard such that the sums of their digits are equal.

thankyou
Debbie Cummins

Hi Debbie,

Here are some suggestions.

  1. Here are two suggestions for the first problem.

    First solution:

    a) N is a multiple of 3 (because 21 is a multiple of 3) so the sum of its digits must be a multiple of 3.
    b) The units digit in 21*AB is B, so B must be 1.
     
    This leaves only 4 possibilities for N: 1011, 1311, 1611 and 1911. You can try them out.
     

    Second solution:

    Let the digits of N be 1xy1; x is the digit in the 100's place and y is the digit in the ten's place. So, by expanding N using place value notation: N= 1000+100x+10y+1

    By removing the end digits, we are left with the 2 digit number xy Now, x is in the ten's place and y is in the one's place So, this number is 10x+y

    Thus, N/21 = 10x+y
    Hence, (1000+100x+10y+1)/21 = 10x+y
    Multiply both sides by 21:
    1000+100x+10y+1=210x+21y
    Subtract 100x and subtract 10y from both sides:
    1001=110x+11y
    Divide both sides by 11:
    91=10x+y
    Thus, using place value notation: x=9 and y=1
    Therefore, N=1911
    (Check: 1911/21 = 91)

  2. 693 = 3 * 3 * 7 * 11.
    For the product of this and N to be a square, you have to pair off the 7 divisor of 693 to a 7 divisor in N, and the 11 divisor in 693 to a 11 divisor in N.
    So, N = 77 times a square. Try all the squares (starting with 4) that make N a 3-digit number.
     
  3. Let d be any day of the 4 year term, so d is a whole number from 1 to 1461 (1996 is a leap year, so there are 366+365+365+365 days in the four year term) A day in the four year term when both houses meet on a Monday occurs only when:

    d-1 is divisible by 10, 12 and 7 Day 1 is the first Monday, the least common multiple of 10, 12 and 7 gives the second monday. Is there another?
     
  4. 1. Remainder of 7 when N is divided d means N=d*K + 7 where K is a whole number
    2. Remainder of 10 when 5N divided d means 5*N=d*L + 10 where L is a whole number
    So, 5*(d*K+7)=d*L+10
    This, 5*d*K+35=d*L+10
    Subtract 10 from both sides:
    d*K+25=d*L
    Divide both sides by d:
    K+(25/d)=L
    But K and L are whole numbers, so 25/d must be a whole number This forces d=1, 5 or 25 And a remainder of 7 when dividing by d forces d=25
     
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