Consider a number 235, where last digit is the sum of first
two digits i.e. 2 + 3 = 5.
How many such 3-digit numbers are there?
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Answer / guest
There are 45 different 3-digit numbers.
The last digit can not be 0.
If the last digit is 1, the only possible number is 101.
(Note that 011 is not a 3-digit number)
If the last digit is 2, the possible numbers are 202 and
112.
If the last digit is 3, the possible numbers are 303, 213
and 123.
If the last digit is 4, the possible numbers are 404, 314,
224 and 134.
If the last digit is 5, the possible numbers are 505, 415,
325, 235 and 145.
Note the pattern here - If the last digit is 1, there is
only one number. If the last digit is 2, there are two
numbers. If the last digit is 3, there are three numbers.
If the last digit is 4, there are four numbers. If the last
digit is 5, there are five numbers. And so on.....
Thus, total numbers are
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
| Is This Answer Correct ? | 71 Yes | 5 No |
Answer / viral avalani
45 nos.
101 112 123 134 145 156 167 178 189 =9
202 213 224 235 246 257 268 279 =8
303 314 325 336 347 358 369 =7
404 415 426 437 448 459 =6
505 516 527 538 549 =5
606 617 628 639 =4
707 718 729 =3
808 819 =2
909 =1
___
Total =45
| Is This Answer Correct ? | 46 Yes | 3 No |
Answer / pratik
Here a and b can also interchange their position.... so
ans is 45 * 2 = 90.
ex - 101 or 110 , 527 or 572 etc...
Now we are not done yet...
We have to subtract those numbers from 90 in which a and b
are equal coz they are counted twice...
So cases where a and b are equal are - 211,422,633,844.
Therefore, ans = 90-4 = 86.
So ans is 86
| Is This Answer Correct ? | 1 Yes | 15 No |
u use any operation solve this problem 45-45=45
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