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findingmyself
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If x is an integer divisible by 3 and y is a non-negative integer such 27 Aug 2025, 08:00
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A
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75% (hard)

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62% (03:14) correct 38% (03:22) wrong based on 211 sessions

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If x is an integer divisible by 3 and y is a non-negative integer such that 6x - 5y = 45 and |y - 5| < 12, Then how many ordered pairs (x, y) satisfy the given conditions?

A. 1
B. 2
C. 3
D. 4
E. 5
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A
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If x is an integer divisible by 3 and y is a non-negative integer such 27 Aug 2025, 09:10
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Y can take values from 0 to 11
We can also check that only at y=9, x=15 which is divisible by 3
Hence only 1 such pair possible

So answer is A
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If x is an integer divisible by 3 and y is a non-negative integer such 30 Aug 2025, 16:03
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Bunuel can you please help with this question
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If x is an integer divisible by 3 and y is a non-negative integer such Updated on: 31 Aug 2025, 02:02
Originally posted by Bunuel on 31 Aug 2025, 02:01.
Last edited by Bunuel on 31 Aug 2025, 02:02, edited 1 time in total.
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findingmyself
If x is an integer divisible by 3 and y is a non-negative integer such that 6x - 5y = 45 and |y - 5| < 12, Then how many ordered pairs (x, y) satisfy the given conditions?

A. 1
B. 2
C. 3
D. 4
E. 5
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|y - 5| < 12

-12 < y - 5 < 12

-7 < y < 17

Since y is a non-negative integer,

0 ≤ y < 17

Next, moving to 6x - 5y = 45. Both 6x and 45 are multiples of 9 (since x is a multiple of 3), thus 5y must also be a multiple of 9, which means y itself must be a multiple of 9. The only multiples of 9 in the above range are 0 and 9.

If y = 0, x is not an integer, so this option is out.
If y = 9, x = 15. That's a valid pair.

So, only one pair of (x, y) exists: (15, 9).

Answer: A.
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If x is an integer divisible by 3 and y is a non-negative integer such 31 Aug 2025, 05:25
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If x is an integer divisible by 3 and y is a non-negative integer such that 6x - 5y = 45 and |y - 5| < 12, Then how many ordered pairs (x, y) satisfy the given conditions?

y>=0
x = (45+5y)/6 = 3k
45 + 5y = 18k
y = (18k -45)/5 >=0
k>=45/18= 5/2 = 2.5
x = 3k>=3*2.5 = 7.5
x>=7.5

-12 < y-5 < 12
-7 < y < 17
0<=y<17

y = 3; x = 10; Not feasible since x is divisible by 3
y = 9; x = 15; Feasible
y = 15; x = 20; Not feasible since x is divisible by 3

IMO A
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