Question

If 1729=1*A+ 7*B+2*C+9*D then how many ones are there inA+B+C+D​

Answer 1

Answer:

Step-by-step explanation:

The possibilities can be listed down below as:

Fix a=1 (Further, fix b for a=1 and find corresponding c and d)

(1,1,1,7), (1,1,2,6), (1,1,3,5), (1,1,4,4), (1,1,5,3), (1,1,6,2), (1,1,7,1);

(1,2,1,6), (1,2,2,5), (1,2,3,4), (1,2,4,3), (1,2,5,2), (1,2,6,1);

(1,3,1,5), (1,3,2,4), (1,3,3,3), (1,3,4,2), (1,3,5,1);

(1,4,1,4), (1,4,2,3), (1,4,3,2), (1,4,4,1);

(1,5,1,3), (1,5,2,2), (1,5,3,1);

(1,6,1,2), (1,6,2,1);

(1,7,1,1);

Fix a=2 (Further, fix b for a=2 and find corresponding c and d)

(2,1,1,6), (2,1,2,5), (2,1,3,4), (2,1,4,3), (2,1,5,2), (2,1,6,1);

(2,2,1,5), (2,2,2,4), (2,2,3,3), (2,2,4,2), (2,2,5,1);

(2,3,1,4), (2,3,2,3), (2,3,3,2), (2,3,4,1);

(2,4,1,3), (2,4,2,2), (2,4,3,1);

(2,5,1,2), (2,5,2,1);

(2,6,1,1);

(WE CAN FIND A PATTERN AT THIS STAGE ONLY, BUT LET US PROCEED FURTHER AND STUDY THE PATTERN AT THE END)

Fix a=3 (Further, fix b for a=3 and find corresponding c and d)

(3,1,1,5), (3,1,2,4), (3,1,3,3), (3,1,4,2), (3,1,5,1);

(3,2,1,4), (3,2,2,3), (3,2,3,2), (3,2,4,1);

(3,3,1,3), (3,3,2,2), (3,3,3,1);

(3,4,1,2), (3,4,2,1);

(3,5,1,1);

Fix a=4 (Further, fix b for a=4 and find corresponding c and d)

(4,1,1,4), (4,1,2,3), (4,1,3,2), (4,1,4,1);

(4,2,1,3), (4,2,2,2), (4,2,3,1);

(4,3,1,2), (4,3,2,1);

(4,4,1,1);

Fix a=5 (Further, fix b for a=5 and find corresponding c and d

(5,1,1,3), (5,1,2,2), (5,1,3,1);

(5,2,1,2), (5,2,2,1);

(5,3,1,1);

Fix a=6 (Further, fix b for a=6 and find corresponding c and d)

(6,1,1,2), (6,1,2,1);

(6,2,1,1);

Fix a=7 (Further, fix b for a=7 and find corresponding c and d)

(7,1,1,1)

The above 4–tuples are the possibilities for (a,b,c,d).

Now let us try to study the pattern of the number of these 4–tuples for various 1st coordinates:

Let us see how many values a can take.

a can be 1 (because this will lead to b+c+d = 9 which is very much possible with b, c, d being positive integers).

a can also be 2 (because this will lead to b+c+d = 8 which is also possible with b, c, d being positive integers).

We can go on similarly till a=7 (because this will lead to b+c+d = 3 which is possible only if b, c, d are all equal to 1).

a cannot take a value greater than 7 because then b+c+d <3 which is not possible when b, c, d are positive integers.

So we have to study the pattern for the various 4–tuples where a can vary from 1 to 7.

For a=1, we have 7 lines. Line 1 has 7 possibilities, Line 2 has 6, Line 3 has 5, and so on till Line 7 which has only 1 possibility.

This gives us 7+6+5+4+3+2+1 (=28) possibilities.

For a=2, we have 6 lines. Line 1 has 6 possibilities, Line 2 has 5, Line 3 has 4, and so on till Line 6 which has only 1 possibility.

This gives us 6+5+4+3+2+1 (=21) possibilities.

For a=3, we have 5 lines. Line 1 has 5 possibilities, Line 2 has 4, Line 3 has 3, and so on till Line 5 which has only 1 possibility.

This gives us 5+4+3+2+1 (=15) possibilities.

Again we can see a relation between fixed value of a and the corresponding possibilities.

So

For a=4, we have 4+3+2+1 (=10) possibilities.

For a=5, we have 3+2+1 (=6) possibilities.

For a=6, we have 2+1 (=3) possibilities.

And finally

For a=7, we have 1 possibility.

Thus, total possibilities

= 28 + 21 + 15 + 10 + 6 + 3 + 1

= 84.

So the answer is 84.

Answer 2

Given :-

• 1729=1*A+ 7*B+2*C+9*D

To Find :-

• how many ones are there in A+B+C+D = ?

Solution :-

→ 1729 = 1*A + 7*B + 2*C + 9*D

→ 1*1000 + 7*100 + 2*10 + 9*1 = 1*A + 7*B + 2*C + 9*D

Comparing we get :-

→ A = 1000

→ B = 100

→ C = 10

→ D = 1

So,

→ A + B + C + D = 1000 + 100 + 10 + 1

→ A + B + C + D = 1111

Hence we can say That, Their are 4 ones in (A + B + C + D).