This topic gives an overview of;
You have studied various types of numbers such as natural numbers, whole numbers, integers and rational numbers. You have also studied a number of interesting properties about them. In previous classes, we explored finding factors and multiples and the relationships among them.
In this chapter, we will explore numbers in more detail. These ideas help in justifying tests of divisibility.
Let us take the number 52 and write it as
52 = 50 + 2 = 10 × 5 + 2
Similarly, the number 37 can be written as
37 = 10 × 3 + 7
In general, any two digit number ab made of digits a and b can be written as
ab = 10 × a + b = 10a + b
ba = 10 × b + a = 10b + a
Let us now take number 351. This is a three digit number. It can also be written as
351 = 300 + 50 + 1 = 100 × 3 + 10 × 5 + 1 × 1
497 = 100 × 4 + 10 × 9 + 1 × 7
Similarly, In general, a 3-digit number abc made up of digits a, b and c is written as
abc = 100 × a + 10 × b + 1 × c
= 100a + 10b + c
In the same way,
cab = 100c + 10a + b
bca = 100b + 10c + a and so on.
Minakshi asks Sundaram to think of a 2-digit number, and then to do whatever she asks him to do, to that number. Their conversation is shown in the following figure.
It so happens that Sundaram chose the number 49. So, he got the reversed number 94; then he added these two numbers and got 49 + 94 = 143. Finally he divided this number by 11 and got 143 ÷ 11 = 13, with no remainder. This is just what Minakshi had predicted.
Now, let us see if we can explain Minakshi’s “trick”.
Suppose Sundaram chooses the number ab, which is a short form for the 2-digit number 10a + b. On reversing the digits, he gets the number ba = 10b + a. When he adds the two numbers he gets:
(10a + b) + (10b + a) = 11a + 11b = 11 (a + b).
So, the sum is always a multiple of 11, just as Minakshi had claimed. Observe here that if we divide the sum by y 11, the quotient is a + b, which is exactly the sum of the digits of chosen number ab. Let us see how Sundaram explains Minakshi’s second “trick”.
Suppose he chooses the 2-digit number ab = 10a + b. After reversing the digits, he gets the number ba = 10b + a. Now Minakshi tells him to do a subtraction, the smaller number from the larger one.
(10a + b) – (10b + a) = 10a + b – 10b – a = 9a – 9b = 9(a – b).
(10b + a) – (10a + b) = 9(b – a).
In each case, the resulting number is divisible by 9. So, the remainder is 0. Observe here that if we divide the resulting number (obtained by subtraction), the quotient is a – b or b – a according as a > b or a < b. You may check the same by taking any other two digit numbers.
Think of a 3-digit number, Now make a new number by putting the digits in reverse order, and subtract the smaller number from the larger one. Divide your answer by 99.There will be no remainder!!!
In fact, Minakshi chose the 3-digit number 349. So she got:
Let us see how this trick works.
Let the 3-digit number chosen by Minakshi be abc = 100a + 10b + c. After reversing the order of the digits, she gets the number cba = 100c + 10b + a. On subtraction:
(100a + 10b + c) – (100c + 10b + a) = 100a + 10b + c – 100c – 10b – a
= 99a – 99c = 99(a – c).
(100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a).
In each case, the resulting number is divisible by 99. So the remainder is 0. Observe that quotient is a – c or c – a. You may check the same by taking other 3-digit numbers.
Think of any 3-digit number. Now use this number to form two more 3-digit numbers, like this:if the number you chose is abc, then
In fact, Sundaram had thought of the 3-digit number 237. After doing what Minakshi had asked, he got the numbers 723 and 372. So he did:
Then he divided the resulting number 1332 by 37: 1332 ÷ 37 = 36, with no remainder
Here we have puzzles in which letters take the place of digits in an arithmetic ‘sum’, and the problem is to find out which letter represents which digit; so it is like cracking a code. Here we stick to problems of addition and multiplication
Here are two rules we follow while doing such puzzles.
A rule that we would like to follow is that the puzzle must have just one answer
Example: Find the digits A and B
Solution: This also has two letters A and B whose values are to be found. Since the ones digit of 3 × A is A, it must be that A = 0 or A = 5.
Now look at B. If B = 1, then BA × B3 would at most be equal to 19 × 19; that is, it would at most be equal to 361. But the product here is 57A, which is more than 500. So we cannot have B = 1.
If B = 3, then BA × B3 would be more than 30 × 30; that is, more than 900. But 57A is less than 600. So, B can not be equal to 3. Putting these two facts together, we see that B = 2 only. So the multiplication is either 20 × 23, or 25 × 23. The first possibility fails, since 20 × 23 = 460. But, the second one works out correctly, since 25 × 23 = 575. So the answer is A = 5, B = 2.
Cite this Simulator: