$11.1951^{1952}-1949^{1951}$ The last two digits of the difference are ( ).





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【Analysis】1951 divided by 4 leaves a remainder of $3,3^{1}$ divided by 4 leaves a remainder of $3,3^{2}$ divided by 4 leaves a remainder of 1, the cycle is 2, so $1951^{1952}$ divided by 4 leaves a remainder of 1, 1951 divided by 25 leaves a remainder of 1, so $1951^{1952}$ divided by 25 leaves a remainder of 1, so $1951^{1952}$ divided by 100 leaves a remainder of 1, which means the last two digits of $1951^{1952}$ are $01$; 1949 divided by 4 leaves a remainder of 1, so $1949^{1951}$ divided by 4 leaves a remainder of 1, 1949 divided by 25 leaves a remainder of $24,24^{1}$ divided by 25 leaves a remainder of $24,24^{2}$ divided by 25 leaves a remainder of 1, the cycle is 2, so $1949^{1951}$ divided by 25 leaves a remainder of 24, so $1949^{1951}$ divided by 100 leaves a remainder of 49, which means the last two digits of $1949^{1951}$ are 49; therefore, the last two digits of the original expression are $101-49=52$