If x, a, and b are positive integers such that when x is

What do you mean by "these problems"? Remainder problems or must be true problems?

Anyway...

Remainders:
Theory: compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html

Practice PS:
remainder-101074.html
remainder-problem-92629.html
number-properties-question-from-qr-2nd-edition-ps-96030.html
remainder-when-k-96127.html
ps-0-to-50-inclusive-remainder-76984.html
good-problem-90442.html
remainder-of-89470.html
number-system-60282.html
remainder-problem-88102.html

Practice DS:
remainder-problem-101740.html
remainder-101663.html
ds-gcd-of-numbers-101360.html
data-sufficiency-with-remainder-98529.html
sum-of-remainders-99943.html
ds8-93971.html
need-solution-98567.html
gmat-prep-ds-remainder-96366.html
gmat-prep-ds-93364.html
ds-from-gmatprep-96712.html
remainder-problem-divisible-by-86839.html
gmat-prep-2-remainder-86155.html
remainder-94472.html
remainder-problem-84967.html

COULD or MUST be true questions:
ds-number-theory-101025.html?hilit=must%20true
number-properties-question-101150.html?hilit=must%20true
gmat-club-please-explain-83605.html?hilit=must%20true
must-be-true-101575.html?hilit=must%20true
gmat-prep-question-101282.html?hilit=must%20true
ab-2-c-is-even-101751.html?hilit=must%20true
mgmat-inequalities-101732.html?hilit=must%20true#p788920
division-and-inequalities-87707.html?hilit=could%20true%20following#p666131

Hope it helps.

If x, a, and b are positive integers such that when x is divided by a, the remainder is b and when x is divided by b, the remainder is a−2, then which of the following must be true?

A)a is even
b)x+b is divisible by a
c)x−1 is divisible by a
d)b=a−1
e)a+2=b+1

If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a-2\), then which of the following must be true?

A. \(a\) is even
B. \(x+b\) is divisible by \(a\)
C. \(x-1\) is divisible by \(a\)
D. \(b=a-1\)
E. \(a+2=b+1\)

When \(x\) is divided by \(a\), the remainder is \(b\) --> \(x=aq+b\) --> \(remainder=b<a=divisor\) (remainder must be less than divisor);
When \(x\) is divided by \(b\), the remainder is \(a-2\) --> \(x=bp+(a-2)\) --> \(remainder=(a-2)<b=divisor\).

So we have that: \(a-2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a-1\) (there is only one integer between \(a-2\) and \(a\), which is \(a-1\) and we are told that this integer is \(b\), hence \(b=a-1\)).

Answer: D.

If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a-2\), then which of the following must be true?

A. \(a\) is even
B. \(x+b\) is divisible by \(a\)
C. \(x-1\) is divisible by \(a\)
D. \(b=a-1\)
E. \(a+2=b+1\)

When \(x\) is divided by \(a\), the remainder is \(b\) --> \(x=aq+b\) --> \(remainder=b<a=divisor\) (remainder must be less than divisor);
When \(x\) is divided by \(b\), the remainder is \(a-2\) --> \(x=bp+(a-2)\) --> \(remainder=(a-2)<b=divisor\).

So we have that: \(a-2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a-1\) (there is only one integer between \(a-2\) and \(a\), which is \(a-1\) and we are told that this integer is \(b\), hence \(b=a-1\)).

Answer: D.

Hi Bunuel,

I solved this problem with a bit different approach
x = p*a + b..........eqn(1)
and x = q*b + (a-2)..............eqn(2)

now, equating eqn(1) and eqn(2)
p*a + b = q*b + (a-2)

a(p-1) = b(q-1) - 2
if we put p = q = 3

we get, 2a = 2b - 2
or a = b - 1
or a + 2 = b + 1 which is option E

would pl tell me where am i wrong with my approach??
Thanks.