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Question 21

Context: Observe this 3×33 \times 3 grid. It is filled following a simple rule — use numbers from 191 - 9 without repeating any of them. There are circled numbers outside the grid. The numbers in the yellow circles are the sums of the corresponding rows and columns.

Q. Fill the grids below based on the rule mentioned above:

Question diagram 1
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Solution

IT-21

Chapter: NUMBER PLAY
Class: 7 (Class 7)
Category: in_text


Question

Context: Observe this 3×33 \times 3 grid. It is filled following a simple rule — use numbers from 191 - 9 without repeating any of them. There are circled numbers outside the grid. The numbers in the yellow circles are the sums of the corresponding rows and columns.

Q. Fill the grids below based on the rule mentioned above:

Question diagram(s):

Question diagram


We need to fill two 3×33 \times 3 grids. Each grid uses numbers from 1 to 9 exactly once. The numbers outside the grid are sums of rows and columns.

Step 1 — Filling the first grid

Let us label the cells of the first grid as RxCyR_x C_y. Here xx is the row number and yy is the column number. We are given some numbers already filled in the grid. The number in cell R1C1R_1 C_1 is 9. The number in cell R3C3R_3 C_3 is 5.

The sum of numbers in Row 1 is 13. R1C1+R1C2+R1C3=13R_1 C_1 + R_1 C_2 + R_1 C_3 = 13 9+R1C2+R1C3=139 + R_1 C_2 + R_1 C_3 = 13 So, the sum of R1C2R_1 C_2 and R1C3R_1 C_3 is: R1C2+R1C3=139R_1 C_2 + R_1 C_3 = 13 - 9 R1C2+R1C3=4R_1 C_2 + R_1 C_3 = \mathbf{4}

The sum of numbers in Column 3 is 12. R1C3+R2C3+R3C3=12R_1 C_3 + R_2 C_3 + R_3 C_3 = 12 R1C3+R2C3+5=12R_1 C_3 + R_2 C_3 + 5 = 12 So, the sum of R1C3R_1 C_3 and R2C3R_2 C_3 is: R1C3+R2C3=125R_1 C_3 + R_2 C_3 = 12 - 5 R1C3+R2C3=7R_1 C_3 + R_2 C_3 = \mathbf{7}

The sum of numbers in Row 3 is 18. R3C1+R3C2+R3C3=18R_3 C_1 + R_3 C_2 + R_3 C_3 = 18 R3C1+R3C2+5=18R_3 C_1 + R_3 C_2 + 5 = 18 So, the sum of R3C1R_3 C_1 and R3C2R_3 C_2 is: R3C1+R3C2=185R_3 C_1 + R_3 C_2 = 18 - 5 R3C1+R3C2=13R_3 C_1 + R_3 C_2 = \mathbf{13}

The sum of numbers in Column 1 is 24. R1C1+R2C1+R3C1=24R_1 C_1 + R_2 C_1 + R_3 C_1 = 24 9+R2C1+R3C1=249 + R_2 C_1 + R_3 C_1 = 24 So, the sum of R2C1R_2 C_1 and R3C1R_3 C_1 is: R2C1+R3C1=249R_2 C_1 + R_3 C_1 = 24 - 9 R2C1+R3C1=15R_2 C_1 + R_3 C_1 = \mathbf{15}

The numbers already used are 9 and 5. The remaining numbers are {1, 2, 3, 4, 6, 7, 8}. From R1C2+R1C3=4R_1 C_2 + R_1 C_3 = 4, the only possible pair is (1, 3). Let us try R1C2=1R_1 C_2 = 1 and R1C3=3R_1 C_3 = 3. Numbers used: 9, 5, 1, 3. Remaining: {2, 4, 6, 7, 8}.

Now we use R1C3+R2C3=7R_1 C_3 + R_2 C_3 = 7. 3+R2C3=73 + R_2 C_3 = 7 R2C3=73R_2 C_3 = 7 - 3 R2C3=4R_2 C_3 = \mathbf{4} Numbers used: 9, 5, 1, 3, 4. Remaining: {2, 6, 7, 8}.

The sum of numbers in Row 2 is 14. R2C1+R2C2+R2C3=14R_2 C_1 + R_2 C_2 + R_2 C_3 = 14 R2C1+R2C2+4=14R_2 C_1 + R_2 C_2 + 4 = 14 So, the sum of R2C1R_2 C_1 and R2C2R_2 C_2 is: R2C1+R2C2=144R_2 C_1 + R_2 C_2 = 14 - 4 R2C1+R2C2=10R_2 C_1 + R_2 C_2 = \mathbf{10} From the remaining numbers {2, 6, 7, 8}, the possible pairs for (R2C1R_2 C_1, R2C2R_2 C_2) that sum to 10 are (2, 8) or (8, 2).

Let us try R2C1=2R_2 C_1 = 2 and R2C2=8R_2 C_2 = 8. Numbers used: 9, 5, 1, 3, 4, 2, 8. Remaining: {6, 7}. Now we use R2C1+R3C1=15R_2 C_1 + R_3 C_1 = 15. 2+R3C1=152 + R_3 C_1 = 15 R3C1=152R_3 C_1 = 15 - 2 R3C1=13R_3 C_1 = \mathbf{13} This number is not allowed as it is greater than 9. So this choice was incorrect.

Let us try R2C1=8R_2 C_1 = 8 and R2C2=2R_2 C_2 = 2. Numbers used: 9, 5, 1, 3, 4, 8, 2. Remaining: {6, 7}. Now we use R2C1+R3C1=15R_2 C_1 + R_3 C_1 = 15. 8+R3C1=158 + R_3 C_1 = 15 R3C1=158R_3 C_1 = 15 - 8 R3C1=7R_3 C_1 = \mathbf{7} Numbers used: 9, 5, 1, 3, 4, 8, 2, 7. Remaining: {6}.

Now we use R3C1+R3C2=13R_3 C_1 + R_3 C_2 = 13. 7+R3C2=137 + R_3 C_2 = 13 R3C2=137R_3 C_2 = 13 - 7 R3C2=6R_3 C_2 = \mathbf{6} All numbers from 1 to 9 are now used: {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Let us check the last sum, Column 2, which is 9. R1C2+R2C2+R3C2=1+2+6R_1 C_2 + R_2 C_2 + R_3 C_2 = 1 + 2 + 6 =9= 9 This matches the given sum. So the grid is filled correctly.

The filled first grid is:

913824765\boxed{\begin{array}{|c|c|c|} \hline 9 & 1 & 3 \\ \hline 8 & 2 & 4 \\ \hline 7 & 6 & 5 \\ \hline \end{array}}

Diagram 1

Step 2 — Filling the second grid

Let us label the cells of the second grid as RxCyR_x C_y. We are given some numbers already filled in the grid. The number in cell R2C1R_2 C_1 is 4. The number in cell R3C3R_3 C_3 is 3.

The sum of numbers in Row 3 is 6. R3C1+R3C2+R3C3=6R_3 C_1 + R_3 C_2 + R_3 C_3 = 6 R3C1+R3C2+3=6R_3 C_1 + R_3 C_2 + 3 = 6 So, the sum of R3C1R_3 C_1 and R3C2R_3 C_2 is: R3C1+R3C2=63R_3 C_1 + R_3 C_2 = 6 - 3 R3C1+R3C2=3R_3 C_1 + R_3 C_2 = \mathbf{3}

The sum of numbers in Column 1 is 12. R1C1+R2C1+R3C1=12R_1 C_1 + R_2 C_1 + R_3 C_1 = 12 R1C1+4+R3C1=12R_1 C_1 + 4 + R_3 C_1 = 12 So, the sum of R1C1R_1 C_1 and R3C1R_3 C_1 is: R1C1+R3C1=124R_1 C_1 + R_3 C_1 = 12 - 4 R1C1+R3C1=8R_1 C_1 + R_3 C_1 = \mathbf{8}

The sum of numbers in Row 2 is 15. R2C1+R2C2+R2C3=15R_2 C_1 + R_2 C_2 + R_2 C_3 = 15 4+R2C2+R2C3=154 + R_2 C_2 + R_2 C_3 = 15 So, the sum of R2C2R_2 C_2 and R2C3R_2 C_3 is: R2C2+R2C3=154R_2 C_2 + R_2 C_3 = 15 - 4 R2C2+R2C3=11R_2 C_2 + R_2 C_3 = \mathbf{11}

The sum of numbers in Column 3 is 17. R1C3+R2C3+R3C3=17R_1 C_3 + R_2 C_3 + R_3 C_3 = 17 R1C3+R2C3+3=17R_1 C_3 + R_2 C_3 + 3 = 17 So, the sum of R1C3R_1 C_3 and R2C3R_2 C_3 is: R1C3+R2C3=173R_1 C_3 + R_2 C_3 = 17 - 3 R1C3+R2C3=14R_1 C_3 + R_2 C_3 = \mathbf{14}

The numbers already used are 4 and 3. The remaining numbers are {1, 2, 5, 6, 7, 8, 9}. From R3C1+R3C2=3R_3 C_1 + R_3 C_2 = 3, the only possible pair is (1, 2). Let us try R3C1=1R_3 C_1 = 1 and R3C2=2R_3 C_2 = 2. Numbers used: 4, 3, 1, 2. Remaining: {5, 6, 7, 8, 9}.

Now we use R1C1+R3C1=8R_1 C_1 + R_3 C_1 = 8. R1C1+1=8R_1 C_1 + 1 = 8 R1C1=81R_1 C_1 = 8 - 1 R1C1=7R_1 C_1 = \mathbf{7} Numbers used: 4, 3, 1, 2, 7. Remaining: {5, 6, 8, 9}.

The sum of numbers in Row 1 is 24. R1C1+R1C2+R1C3=24R_1 C_1 + R_1 C_2 + R_1 C_3 = 24 7+R1C2+R1C3=247 + R_1 C_2 + R_1 C_3 = 24 So, the sum of R1C2R_1 C_2 and R1C3R_1 C_3 is: R1C2+R1C3=247R_1 C_2 + R_1 C_3 = 24 - 7 R1C2+R1C3=17R_1 C_2 + R_1 C_3 = \mathbf{17} From the remaining numbers {5, 6, 8, 9}, the only possible pairs for (R1C2R_1 C_2, R1C3R_1 C_3) that sum to 17 are (8, 9) or (9, 8).

Let us try R1C2=9R_1 C_2 = 9 and R1C3=8R_1 C_3 = 8. Numbers used: 4, 3, 1, 2, 7, 9, 8. Remaining: {5, 6}.

Now we use R2C2+R2C3=11R_2 C_2 + R_2 C_3 = 11. From the remaining numbers {5, 6}, the possible pairs for (R2C2R_2 C_2, R2C3R_2 C_3) that sum to 11 are (5, 6) or (6, 5).

Let us try R2C2=5R_2 C_2 = 5 and R2C3=6R_2 C_3 = 6. All numbers from 1 to 9 are now used: {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Let us check the remaining sums. Column 2 sum is 16. R1C2+R2C2+R3C2=9+5+2R_1 C_2 + R_2 C_2 + R_3 C_2 = 9 + 5 + 2 =16= 16 This matches the given sum.

Column 3 sum is 17. R1C3+R2C3+R3C3=8+6+3R_1 C_3 + R_2 C_3 + R_3 C_3 = 8 + 6 + 3 =17= 17 This matches the given sum. So the grid is filled correctly.

The filled second grid is:

798456123\boxed{\begin{array}{|c|c|c|} \hline 7 & 9 & 8 \\ \hline 4 & 5 & 6 \\ \hline 1 & 2 & 3 \\ \hline \end{array}}

Diagram 2

Answer

(i) The first grid is: 913824765\begin{array}{|c|c|c|} \hline 9 & 1 & 3 \\ \hline 8 & 2 & 4 \\ \hline 7 & 6 & 5 \\ \hline \end{array} (ii) The second grid is: 798456123\begin{array}{|c|c|c|} \hline 7 & 9 & 8 \\ \hline 4 & 5 & 6 \\ \hline 1 & 2 & 3 \\ \hline \end{array}

More questions in IT

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What do the numbers in the figure below tell us?

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Q2

What do you think these numbers mean?

The children rearrange themselves and each one says a number based on the new arrangement.

Q3

Context: The children rearrange themselves and each one says a number based on the new arrangement.

Q. Could you figure out what these numbers convey? Observe and try to find out.

Q4

Write down the number each child should say based on this rule for the arrangement shown below.

Q5

Kishor has some number cards and is working on a puzzle: There are 5 boxes, and each box should contain exactly 1 number card. The numbers in the boxes should sum to 30. Can you help him find a way to do it?

Can you figure out which 5 cards add to 30? Is it possible? There are many ways of choosing 5 cards from this collection. Is there a way to find a solution without checking all possibilities? Let us find out.

Q6

Context: Kishor has some number cards and is working on a puzzle: There are 5 boxes, and each box should contain exactly 1 number card. The numbers in the boxes should sum to 30. Can you help him find a way to do it? Can you figure out which 5 cards add to 30? Is it possible? There are many ways of choosing 5 cards from this collection. Is there a way to find a solution without checking all possibilities? Let us find out.

Q. Add a few even numbers together. What kind of number do you get? Does it matter how many numbers are added?

Q7

Context: As we see in the figure, adding any number of even numbers will result in a number which can still be arranged in pairs without any leftovers. In other words, the sum will always be an even number.

Q. Now, add a few odd numbers together. What kind of number do you get? Does it matter how many odd numbers are added?

Q8

Context: Can we also think of an odd number as one less than a collection of pairs? This figure shows that the sum of two odd numbers must always be even! This along with the other figures here are more examples of a proof!

Q. What about adding 3 odd numbers? Can the resulting sum be arranged in pairs?

Q9

Explore what happens to the sum of: (a) 4 odd numbers (b) 5 odd numbers (c) 6 odd numbers

Q10

Two siblings, Martin and Maria, were born exactly one year apart. Today they are celebrating their birthday. Maria exclaims that the sum of their ages is 112. Is this possible? Why or why not?

Q11

Context: Small Squares in Grids In a 3×33 \times 3 grid, there are 9 small squares, which is an odd number. Meanwhile, in a 3×43 \times 4 grid, there are 12 small squares, which is an even number.

Q. Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?

Q12

Find the parity of the number of small squares in these grids:

(a) 27×1327 \times 13

(b) 42×7842 \times 78

(c) 135×654135 \times 654

Q13

Context: Consider the algebraic expression: 3n+43n + 4. For different values of nn, the expression has different parity:

(a) Come up with an expression that always has even parity. Some examples are: 100p100p and 48w248w - 2. Try to find more.

(b) Come up with expressions that always have odd parity.

(c) Come up with other expressions, like 3n+43n + 4, which could have either odd or even parity.

(d) The expression 6k+26k + 2 evaluates to 8,14,20,8, 14, 20, \dots (for k=1,2,3,k = 1, 2, 3, \dots) — many even numbers are missing.

(e) Are there expressions using which we can list all the even numbers? Hint: All even numbers have a factor 2.

(f) Are there expressions using which we can list all odd numbers?

Q19

Context: We saw earlier how to express the nthn^{\text{th}} term of the sequence of multiples of 44, where nn is the letter-number that denotes a position in the sequence (e.g., first, twenty third, hundred and seventeenth, etc.).

(1) What would be the nthn^{\text{th}} term for multiples of 22? Or, what is the nthn^{\text{th}} even number?

Let us consider odd numbers.

(2) What is the 100th odd number?

To answer this question, consider the following question:

(3) What is the 100th even number?

(4) Write a formula to find the nthn^{\text{th}} odd number.

Q21

Context: Observe this 3×33 \times 3 grid. It is filled following a simple rule — use numbers from 191 - 9 without repeating any of them. There are circled numbers outside the grid. The numbers in the yellow circles are the sums of the corresponding rows and columns.

Q. Fill the grids below based on the rule mentioned above:

Q22

Make a couple of questions like this on your own and challenge your peers.

Q27

Can 1 occur in a corner position? For example, can it be placed as follows?

Q. If yes, then there should exist three ways of adding 1 with two other numbers to give 15. We have 1 + 5 + 9 = 1 + 6 + 8 = 15. Is any other combination possible?

Q. Similarly, can 9 can be placed in a corner position?

Q29

Can you find the other possible positions for 1 and 9?

Now, we have one full row or column of the magic square! Try completing it!

[Hint: First fill the row or columns containing 1 and 9]

Q30

Choose any magic square that you have made so far using consecutive numbers. If mm is the letter-number of the number in the centre, express how other numbers are related to mm, how much more or less than mm.

[Hint: Remember, how we described a 2×22 \times 2 grid of a calendar month in the Algebraic Expressions chapter].

Q31

Context: Choose any magic square that you have made so far using consecutive numbers. If mm is the letter-number of the number in the centre, express how other numbers are related to mm, how much more or less than mm.

[Hint: Remember, how we described a 2×22 \times 2 grid of a calendar month in the Algebraic Expressions chapter].

Q. Once the generalised form is obtained, share your observations with the class.

Q32

Chautīsā means 34. Why do you think they called it the Chautīsā Yantra? Every row, column and diagonal in this magic square adds up to 34. Can you find other patterns of four numbers in the square that add up to 34?

Q33

How many rhythms are there with 8 beats consisting of short syllables (1 beat) and long syllables (2 beats)? That is, in how many ways can one fill 8 beats with short and long syllables, where a short syllable takes one beat of time and a long syllable takes two beats of time?

Q34

Context: A short syllable takes one beat of time and a long syllable takes two beats of time. Some possibilities to fill 8 beats are:

  • long long long long
  • short short short short short short short short
  • short long long short long
  • long long short short long

Q. Can you find others?

Q35

Context: We can write the number 8 as a sum of 1's and 2's in several ways, for example:

  • 8 = 2 + 2 + 2 + 2
  • 8 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
  • 8 = 1 + 2 + 2 + 1 + 2
  • 8 = 2 + 2 + 1 + 1 + 2

Q. Do you see other ways?

Q36

Try writing the number 5 as a sum of 1s and 2s in all possible ways in your notebook! How many ways did you find? (You should find 8 different ways!) Can you figure out the answer without listing down all the possibilities? Can you try it for n = 8?

Q37

Context: Thus, there are 8 rhythms having 5 beats! The reason this method works is that every 5-beat rhythm must begin with either a '1+' or a '2+'. If it begins with a '1+', then the remaining numbers must give a 4-beat rhythm, and we can write all those down. If it begins with a 2+, then the remaining number must give a 3-beat rhythm, and we can write all those down. Therefore, the number of 5-beat rhythms is the number of 4-beat rhythms, plus the number of 3-beat rhythms. How many 6-beat rhythms are there? By the same reasoning, it will be the number of 5-beat rhythms plus the number of 4-beat rhythms, i.e., 8+5=138 + 5 = 13. Thus, there are 13 rhythms having 6 beats.

Q. Use the systematic method to write down all 6-beat rhythms, i.e., write 6 as the sum of 1's and 2's in all possible ways. Did you get 13 ways?

Q38

Write the next 3 numbers in the sequence: 1,2,3,5,8,13,21,34,55,89,,,,1, 2, 3, 5, 8, 13, 21, 34, 55, 89, \underline{\quad}, \underline{\quad}, \underline{\quad}, \dots

If you have to write one more number in the sequence above, can you tell whether it will be an odd number or an even number (without adding the two previous numbers)?

Q39

Context: 1,2,3,5,8,13,21,34,55,89,1, 2, 3, 5, 8, 13, 21, 34, 55, 89, \dots

What is the parity of each number in the sequence? Do you notice any pattern in the sequence of parities?

Q40

Context: Let us look at one more example. Here K2\text{K2} means that the number is a 2-digit number having the digit '2' in the units place and 'K' in the tens place. K2\text{K2} is added to itself to give a 3-digit sum HMM\text{HMM}:

K2+ K2HMM\begin{array}{r} \text{K2} \\ +\ \text{K2} \\ \hline \text{HMM} \end{array}

Q. What digit should the letter M\text{M} correspond to? Both the tens place and the units place of the sum have the same digit. What about H\text{H}? Can it be 2? Can it be 3?

Q41

Context: These types of questions can be interesting and fun to solve! Here are some more questions like this for you to try out. Find out what each letter stands for. Share how you thought about each question with your classmates; you may find some new approaches.

Q. Find out what each letter stands for:

(i) YY+ZZOO\begin{array}{r} \text{YY} \\ +\quad \text{Z} \\ \hline \text{ZOO} \end{array}

(ii) B5+3DED5\begin{array}{r} \text{B5} \\ +\quad \text{3D} \\ \hline \text{ED5} \end{array}

(iii) KP+KPPRR\begin{array}{r} \text{KP} \\ +\quad \text{KP} \\ \hline \text{PRR} \end{array}

(iv) C1+C1FF\begin{array}{r} \text{C1} \\ +\quad \text{C} \\ \hline \text{1FF} \end{array}

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