I am taking from The Official Hunter College High School Test: problem 76 on page 20. We read the following.
In the expression below, each letter represents a one digit number. Where the same letter appears, it represents the same number in each case. Each distinct letter represents a
different number. In order to make the equation true, what number must replace C?
A great start is to decode each AAA, AAB, and ABC. It helps to look at this problem wholly; particularly we look at the leading sum on the left wall (of the same types). We glean that either: (1) A + A + A = 20, (2) A + A + A + 1 = 20 or (3) A + A + A + 2 = 20: its very important to remember that given three numbers each less than ten, the sum of them which is great, is at most 2 in the tens place. This means that each row can only donate a 1 or 2 to the next. We can conclude that our line is...
Maths:Number System Concepts:
42 = 7*2*3
Here 7,2 and 3 are factors of 42.
42 is a multiple of 6.
42 is a multiple of 7.
If a, b are integers(b is not 0)and a/b is an integer.
=> b is a factor of a.
=>a is divisible by b.
=>a is a mulitple of b.
E.g. Positive factors of 12 = 1,2,3,4,6,12.
E.g. Multiples of 12 = 12,24,36,0, -12,-36.......
To find total number of factors.
180= 2^2 *3^2 *5
=>18*2= 36 ;to include negative factors
Always consider intensive counting unless mentioned otherwise.
Never forget to check negative numbers.
Numbers with exactly three divisors:
Squares of prime numbers.
E.g. => 49,121
They have exactly three factors : 1,7,49 & 1,11,121.
I have discovered a new way to count even more than ten by using those same ten fingers you can use each section of one finger to count each line on that finger plus the finger in which on all of them have three lines. So imagine now you can count up to three times more (30). I just wanted to shared this since a student I tutor from India counted this way and I was amazed.