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29 Jan 2023, 03:47
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mysterymanrog
took a while! This was my approach: 10000000000-(ab) Now, using subtraction property we get that the sum of digits is (under the assumption that b!=0): 9*8+(9-a)+(10-b) we need it to be equal to 84, so 9*8+(9-a)+(10-b)=84 (9-a)+(10-b)=12 now we can just have fun plugging in numbers a=1,b=6 a=2,b=5 a=4,b=3 a=5,b=2 a=6,b=1 a=7,b=0 -> this is left out, because we have assumed b not equal to 0 The case of b=0: we have 9*9+(10-a)=84 (10-a)=3 a=7,b=0 total of 6 values of Z. Choice D.
Any other way to solve this? 
Updated on: 29 Jan 2023, 03:48
Originally posted by mysterymanrog on 29 Jan 2023, 03:47.
Last edited by mysterymanrog on 29 Jan 2023, 03:48, edited 1 time in total.
Last edited by mysterymanrog on 29 Jan 2023, 03:48, edited 1 time in total.
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Probably
there is 100% some type of pattern there, which would expedite the process
there is 100% some type of pattern there, which would expedite the process
29 Jan 2023, 04:01
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Rashiudhwani
Any other way to solve this?
The trick is to focus only on the last two digits. mysterymanrog did a great job already !
Here are my two cents on this -
Once you’ve figured out that all numbers except the last two yield 9 and thus contribute 72 to the total. The sum of the last two digits should be 12.
If you focus on the last two digits only as shown, after writing the first few (I have written the first three) terms you will see a pattern. Just follow that pattern and stop once you’ve reached 9 in the tens place. Of course, we can’t have 9 in the tens place, so discard that case.
There are 6 numbers possible.
